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630090, Россия, Новосибирск, пр-т Ак. Лаврентьева, 5


1. Основные понятия химической термодинамики

При решении этих задач очень полезно использовать метод якобианов, заключающийся в том, что частные производные, требующие преобразования, переводят в якобиан по очевидному соотношению:

( ∂u ∂x ) y = ∂( u,y ) ∂( x,y ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadwhaaeaacqGHciITcaWG4baaaaGaayjkaiaawMcaamaaBaaaleaacaWG5baabeaakiabg2da9maalaaabaGaeyOaIy7aaeWaaeaacaWG1bGaaiilaiaadMhaaiaawIcacaGLPaaaaeaacqGHciITdaqadaqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaaaaaaaa@49C4@ ,

(1)

а затем преобразуют якобиан, используя следующие три алгебраические тождества:

∂( u,v ) ∂( x,y ) =− ∂( v,u ) ∂( x,y ) =− ∂( u,v ) ∂( y,x ) = ∂( v,u ) ∂( y,x ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@67D1@ ,

(2)

( ∂( u,v ) ∂( x,y ) ) −1 = ∂( x,y ) ∂( u,v ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2oaabmaabaGaamyDaiaacYcacaWG2baacaGLOaGaayzkaaaabaGaeyOaIy7aaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaeyypa0ZaaSaaaeaacqGHciITdaqadaqaaiaadIhacaGGSaGaamyEaaGaayjkaiaawMcaaaqaaiabgkGi2oaabmaabaGaamyDaiaacYcacaWG2baacaGLOaGaayzkaaaaaaaa@50D7@ ,

(3)

∂( u,v ) ∂( x,y ) ∂( x,y ) ∂( w,z ) = ∂( u,v ) ∂( w,z ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITdaqadaqaaiaadwhacaGGSaGaamODaaGaayjkaiaawMcaaaqaaiabgkGi2oaabmaabaGaamiEaiaacYcacaWG5baacaGLOaGaayzkaaaaamaalaaabaGaeyOaIy7aaeWaaeaacaWG4bGaaiilaiaadMhaaiaawIcacaGLPaaaaeaacqGHciITdaqadaqaaiaadEhacaGGSaGaamOEaaGaayjkaiaawMcaaaaacqGH9aqpdaWcaaqaaiabgkGi2oaabmaabaGaamyDaiaacYcacaWG2baacaGLOaGaayzkaaaabaGaeyOaIy7aaeWaaeaacaWG3bGaaiilaiaadQhaaiaawIcacaGLPaaaaaaaaa@58B3@ .

(4)

Отметим также, что все четыре соотношения Максвелла тождественны одному единственному соотношению, записанному через якобиан:

∂( T,S ) ∂( P,V ) =1. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITdaqadaqaaiaadsfacaGGSaGaam4uaaGaayjkaiaawMcaaaqaaiabgkGi2oaabmaabaGaamiuaiaacYcacaWGwbaacaGLOaGaayzkaaaaaiabg2da9iaaigdacaGGUaaaaa@430F@

(5

Приведенный выше метод и будет использован ниже при решении задач.

Кроме того, полезно использовать алгебраическое соотношение между частными производными:

( ∂x ∂y ) z ( ∂y ∂z ) x ( ∂z ∂x ) y =−1 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadIhaaeaacqGHciITcaWG5baaaaGaayjkaiaawMcaamaaBaaaleaacaWG6baabeaakmaabmaabaWaaSaaaeaacqGHciITcaWG5baabaGaeyOaIyRaamOEaaaaaiaawIcacaGLPaaadaWgaaWcbaGaamiEaaqabaGcdaqadaqaamaalaaabaGaeyOaIyRaamOEaaqaaiabgkGi2kaadIhaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadMhaaeqaaOGaeyypa0JaeyOeI0IaaGymaaaa@4F5A@

(6)

и не забывать Второе начало термодинамики и определения теплоемкостей:

c V =T ( ∂S ∂T ) V ,  c P =T ( ∂S ∂T ) P MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaaBaaaleaacaWGwbaabeaakiabg2da9iaadsfadaqadaqaamaalaaabaGaeyOaIyRaam4uaaqaaiabgkGi2kaadsfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadAfaaeqaaOGaaiilaiaaywW7caWGJbWaaSbaaSqaaiaadcfaaeqaaOGaeyypa0JaamivamaabmaabaWaaSaaaeaacqGHciITcaWGtbaabaGaeyOaIyRaamivaaaaaiaawIcacaGLPaaadaWgaaWcbaGaamiuaaqabaaaaa@4E13@

(7)

10. (1/Э-06).* Известно термическое уравнение состояния газа Ван-дер-Ваальса: (P+ a V 2 )(V−b)=RT. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadcfacqGHRaWkdaWcaaqaaiaadggaaeaacaWGwbWaaWbaaSqabeaacaaIYaaaaaaakiaacMcacaGGOaGaamOvaiabgkHiTiaadkgacaGGPaGaeyypa0JaamOuaiaadsfacaGGUaaaaa@4331@ Выведите калорическое уравнение состояния газа Ван-дер-Ваальса U = U(T,V).

Решение. В дифференциальной форме калорическое уравнение состояния записывается как dU= ( ∂U ∂V ) T dV+ ( ∂U ∂T ) v dT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwfacqGH9aqpdaqadaqaamaalaaabaGaeyOaIyRaamyvaaqaaiabgkGi2kaadAfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadsfaaeqaaOGaamizaiaadAfacqGHRaWkdaqadaqaamaalaaabaGaeyOaIyRaamyvaaqaaiabgkGi2kaadsfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadAhaaeqaaOGaamizaiaadsfaaaa@4B90@ .

Из Второго начала dU=TdS−PdV MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwfacqGH9aqpcaWGubGaamizaiaadofacqGHsislcaWGqbGaamizaiaadAfaaaa@3ED6@ следует, что ( ∂U ∂V ) T =T ( ∂S ∂V ) T −P MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadwfaaeaacqGHciITcaWGwbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGubaabeaakiabg2da9iaadsfadaqadaqaamaalaaabaGaeyOaIyRaam4uaaqaaiabgkGi2kaadAfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadsfaaeqaaOGaeyOeI0Iaamiuaaaa@47DE@ и ( ∂U ∂T ) V =T ( ∂S ∂T ) V = c V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadwfaaeaacqGHciITcaWGubaaaaGaayjkaiaawMcaamaaBaaaleaacaWGwbaabeaakiabg2da9iaadsfadaqadaqaamaalaaabaGaeyOaIyRaam4uaaqaaiabgkGi2kaadsfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadAfaaeqaaOGaeyypa0Jaam4yamaaBaaaleaacaWGwbaabeaaaaa@4911@

( ∂S ∂V ) T = ∂(S,T) ∂(V,T) = ∂(S,T) ∂(V,T) ( ∂(V,P) ∂(S,T) )= ∂(V,P) ∂(V,T) = ( ∂P ∂T ) V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7316@ .

Таким образом, dU=( T ( ∂P ∂T ) V −P )dV+ c V dT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwfacqGH9aqpdaqadaqaaiaadsfadaqadaqaamaalaaabaGaeyOaIyRaamiuaaqaaiabgkGi2kaadsfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadAfaaeqaaOGaeyOeI0IaamiuaaGaayjkaiaawMcaaiaadsgacaWGwbGaey4kaSIaam4yamaaBaaaleaacaWGwbaabeaakiaadsgacaWGubaaaa@4A5F@ (для любого газа).

Для идеального газа множитель при dV равен нулю. Для газа Ван-дер-Ваальса ( ∂P ∂T ) V = ( ∂( RT ( V−b ) − a V 2 ) ∂T ) V = R ( V−b ) = P+ a V 2 T MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5C27@ и, следовательно, dU= a V 2 dV+ c V dT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwfacqGH9aqpdaWcaaqaaiaadggaaeaacaWGwbWaaWbaaSqabeaacaaIYaaaaaaakiaadsgacaWGwbGaey4kaSIaam4yamaaBaaaleaacaWGwbaabeaakiaadsgacaWGubaaaa@41DB@ .

Требуемое калорическое уравнение получаем интегрированием dU.

16. (1/Э-05).* Углекислый газ подчиняется уравнению состояния Ван-дер-Ваальса с параметрами a = 0,364 Дж.м3.моль–2 и b = 4, 27.10–5 м3/моль. Оцените изменение внутренней энергии в процессе сжатия одного моля CO2 с объема V1 = 10 л до V2 = 1 л, проводимом при 298 К:

Решение:В дифференциальной форме калорическое уравнение состояния записывается как (см. решение выше) ΔU= ∫ V 2 V 2 ( ∂U ∂V ) T dV = ∫ V 1 V 2 ( T ( ∂P ∂T ) V −P )dV MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5C33@ .

Для газа Ван-дер-Ваальса (см. решение выше)

ΔU= ∫ V 1 V 2 a V 2 dV =−a( 1 V 2 − 1 V 1 ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaamyvaiabg2da9maapehabaWaaSaaaeaacaWGHbaabaGaamOvamaaCaaaleqabaGaaGOmaaaaaaGccaWGKbGaamOvaaWcbaGaamOvamaaBaaameaacaaIXaaabeaaaSqaaiaadAfadaWgaaadbaGaaGOmaaqabaaaniabgUIiYdGccqGH9aqpcqGHsislcaWGHbWaaeWaaeaadaWcaaqaaiaaigdaaeaacaWGwbWaaSbaaSqaaiaaikdaaeqaaaaakiabgkHiTmaalaaabaGaaGymaaqaaiaadAfadaWgaaWcbaGaaGymaaqabaaaaaGccaGLOaGaayzkaaaaaa@4E3E@ = –0,364 (10 3 – 10 2) Дж = –330 Дж.

17. (2/1-06).* Доказать соотношение ( ∂T ∂V ) U = p− ( ∂p ∂T ) V T C V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadsfaaeaacqGHciITcaWGwbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGvbaabeaakiabg2da9maalaaabaGaamiCaiabgkHiTmaabmaabaWaaSaaaeaacqGHciITcaWGWbaabaGaeyOaIyRaamivaaaaaiaawIcacaGLPaaadaWgaaWcbaGaamOvaaqabaGccaWGubaabaGaam4qamaaBaaaleaacaWGwbaabeaaaaaaaa@49FA@ . Как будет изменяться при адиабатическом расширении в вакуум температура неидеального газа c фактором сжимаемости PV RT ≡Z(V,T) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGqbGaamOvaaqaaiaadkfacaWGubaaaiabggMi6kaadQfacaGGOaGaamOvaiaacYcacaWGubGaaiykaaaa@3FC2@ ?

Решение. Сначала обсудим, что означает "адиабатическое расширение в вакуум". Расширение в вакуум – это необратимый (и, следовательно, неравновесный) процесс. Поэтому условие адиабатичности ни в коем случае не означает S = const, хотя для равновесного процесса это было бы верно. Поскольку при расширении в вакуум газ не совершает работы, то в соответствии с Первым началом адиабатичность означает постоянство внутренней энергии: U = const. Таким образом, соотношение, которое требуется доказать, и даст ответ на вопрос задачи (на самом деле, в текст задачи просто введена подсказка).

Итак, докажем соотношение: ( ∂T ∂V ) U = ∂(T,U) ∂(V,U) = ∂ (T,U) ∂(V,T) ∂ (V,T) ∂(V,U) = MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadsfaaeaacqGHciITcaWGwbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGvbaabeaakiabg2da9maalaaabaGaeyOaIyRaaiikaiaadsfacaGGSaGaamyvaiaacMcaaeaacqGHciITcaGGOaGaamOvaiaacYcacaWGvbGaaiykaaaacqGH9aqpdaWcaaqaaiabgkGi2kaacIcacaWGubGaaiilaiaadwfacaGGPaWaaSbaaSqaaaqabaGccqGHciITcaGGOaGaamOvaiaacYcacaWGubGaaiykaaqaaiabgkGi2kaacIcacaWGwbGaaiilaiaadsfacaGGPaWaaSbaaSqaaaqabaGccqGHciITcaGGOaGaamOvaiaacYcacaWGvbGaaiykaaaacqGH9aqpaaa@5F86@

< =− ( ∂U ∂V ) T ( ∂T ∂U ) V =−( T ( ∂S ∂V ) T −P ) 1 C V = P− ( ∂P ∂T ) V T C V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@62B8@ . Доказано.

Теперь применим к P= Z(T,V)RT V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2da9maalaaabaGaamOwaiaacIcacaWGubGaaiilaiaadAfacaGGPaGaamOuaiaadsfaaeaacaWGwbaaaaaa@3EFF@

P− ( ∂P ∂T ) V T=P− ZRT V − R T 2 V ( ∂Z ∂T ) V =− R T 2 V ( ∂Z ∂T ) V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5E0C@ .

Ответ: ( ∂T ∂V ) U =− R T 2 V C V ( ∂Z ∂T ) V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadsfaaeaacqGHciITcaWGwbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGvbaabeaakiabg2da9iabgkHiTmaalaaabaGaamOuaiaadsfadaahaaWcbeqaaiaaikdaaaaakeaacaWGwbGaam4qamaaBaaaleaacaWGwbaabeaaaaGcdaqadaqaamaalaaabaGaeyOaIyRaamOwaaqaaiabgkGi2kaadsfaaaaacaGLOaGaayzkaaWaaSbaaSqaaiaadAfaaeqaaaaa@4B94@ .

24. (2/1-04).* Показать, что c p − c V =−T ∂ 2 G ∂T∂P ∂ 2 A ∂T∂V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaaBaaaleaacaWGWbaabeaakiabgkHiTiaadogadaWgaaWcbaGaamOvaaqabaGccqGH9aqpcqGHsislcaWGubWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGhbaabaGaeyOaIyRaamivaiabgkGi2kaadcfaaaWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGbbaabaGaeyOaIyRaamivaiabgkGi2kaadAfaaaaaaa@4D10@

Решение.Cначала упростим выражение:
∂ 2 G ∂T∂P = ( ∂V ∂T ) P , MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGhbaabaGaeyOaIyRaamivaiabgkGi2kaadcfaaaGaeyypa0ZaaeWaaeaadaWcaaqaaiabgkGi2kaadAfaaeaacqGHciITcaWGubaaaaGaayjkaiaawMcaamaaBaaaleaacaWGqbaabeaakiaacYcaaaa@4676@ а ∂ 2 A ∂T∂V =− ( ∂P ∂T ) V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGbbaabaGaeyOaIyRaamivaiabgkGi2kaadAfaaaGaeyypa0JaeyOeI0YaaeWaaeaadaWcaaqaaiabgkGi2kaadcfaaeaacqGHciITcaWGubaaaaGaayjkaiaawMcaamaaBaaaleaacaWGwbaabeaaaaa@46A9@ . Требуется показать, что

c p − c V =T ( ∂V ∂T ) P ( ∂P ∂T ) V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaaBaaaleaacaWGWbaabeaakiabgkHiTiaadogadaWgaaWcbaGaamOvaaqabaGccqGH9aqpcaWGubWaaeWaaeaadaWcaaqaaiabgkGi2kaadAfaaeaacqGHciITcaWGubaaaaGaayjkaiaawMcaamaaBaaaleaacaWGqbaabeaakmaabmaabaWaaSaaaeaacqGHciITcaWGqbaabaGaeyOaIyRaamivaaaaaiaawIcacaGLPaaadaWgaaWcbaGaamOvaaqabaaaaa@4B03@ (это – задача 20).

c p =T ( ∂S ∂T ) P =T ∂(S,P) ∂(T,P) =T ∂(T,V) ∂(T,P) ∂(S,P) ∂(T,V) =T ( ∂V ∂P ) T ∂(S,P) ∂(T,V) . MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@77AA@

∂(S,P) ∂(T,V) = ( ∂S ∂T ) V ( ∂P ∂V ) T − ( ∂P ∂T ) V ( ∂S ∂V ) T = c V T ( ∂P ∂V ) T − ( ∂P ∂T ) V ∂(S,T) ∂(V,T) =     = c V T ( ∂P ∂V ) T − ( ∂P ∂T ) V ∂(V,P) ∂(V,T) = c V T ( ∂P ∂V ) T − ( ∂P ∂T ) V ( ∂P ∂T ) V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@BB1D@

Подставляем: c p = c V −T ( ∂V ∂P ) T ( ∂P ∂T ) V 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaaBaaaleaacaWGWbaabeaakiabg2da9iaadogadaWgaaWcbaGaamOvaaqabaGccqGHsislcaWGubWaaeWaaeaadaWcaaqaaiabgkGi2kaadAfaaeaacqGHciITcaWGqbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGubaabeaakmaabmaabaWaaSaaaeaacqGHciITcaWGqbaabaGaeyOaIyRaamivaaaaaiaawIcacaGLPaaadaqhaaWcbaGaamOvaaqaaiaaikdaaaaaaa@4BC0@ (это – задача 21).

Преобразуем: ( ∂V ∂P ) T ( ∂P ∂T ) V =− ( ∂V ∂T ) P MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiabgkGi2kaadAfaaeaacqGHciITcaWGqbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGubaabeaakmaabmaabaWaaSaaaeaacqGHciITcaWGqbaabaGaeyOaIyRaamivaaaaaiaawIcacaGLPaaadaqhaaWcbaGaamOvaaqaaaaakiabg2da9iabgkHiTmaabmaabaWaaSaaaeaacqGHciITcaWGwbaabaGaeyOaIyRaamivaaaaaiaawIcacaGLPaaadaWgaaWcbaGaamiuaaqabaaaaa@4D43@ и получаем требуемое тождество.

27. (1/1-06).* Обратимые процессы, в ходе которых теплоемкость системы C остаётся постоянной, называют политропными. Найдите зависимость Р(V,T) для политропного процесса (уравнение политропы) для идеального газа. Какие политропные процессы вам известны?

Решение.Из Первого начала δQ=CdT=dU+PdV MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiTdqMaamyuaiabg2da9iaadoeacaWGKbGaamivaiabg2da9iaadsgacaWGvbGaey4kaSIaamiuaiaadsgacaWGwbaaaa@423C@ . По условию газ идеальный: dU= C V dT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwfacqGH9aqpcaWGdbWaaSbaaSqaaiaadAfaaeqaaOGaamizaiaadsfaaaa@3C51@ . Тогда (C− C V )dT=PdV MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadoeacqGHsislcaWGdbWaaSbaaSqaaiaadAfaaeqaaOGaaiykaiaadsgacaWGubGaeyypa0JaamiuaiaadsgacaWGwbaaaa@4035@ .

Из термического уравнения состояния идеального газа следует, что dT= PdV+VdP R MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadsfacqGH9aqpdaWcaaqaaiaadcfacaWGKbGaamOvaiabgUcaRiaadAfacaWGKbGaamiuaaqaaiaadkfaaaaaaa@3FB0@ . Тогда, заменив dT, получим VdP=− C P −C C V −C ⋅PdV. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiaadsgacaWGqbGaeyypa0JaeyOeI0YaaSaaaeaacaWGdbWaaSbaaSqaaiaadcfaaeqaaOGaeyOeI0Iaam4qaaqaaiaadoeadaWgaaWcbaGaamOvaaqabaGccqGHsislcaWGdbaaaiabgwSixlaadcfacaWGKbGaamOvaiaac6caaaa@4734@ Интегрируем и получаем уравнение состояния
PVn = const, где n= C P −C C V −C MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabg2da9maalaaabaGaam4qamaaBaaaleaacaWGqbaabeaakiabgkHiTiaadoeaaeaacaWGdbWaaSbaaSqaaiaadAfaaeqaaOGaeyOeI0Iaam4qaaaaaaa@3F0C@ .

Хорошо известные всем политропы: изобара (n = 0, C=CP); изохора (n = ∞, C=CV); адиабата (n = γ = CP/CV, C=0).

PV= const (изотерма) – это тоже политропный процесс, но теплоемкость в этом случае не имеет смысла (С → ∞).

32. (1/Э-04).* Распространение звука в идеальном газе можно рассматривать как адиабатический процесс. Из гидродинамики известно, что скорость звука с = {(∂P/∂ρ)адиаб}0,5, где P – давление, а ρ – плотность газа. Найти скорость звука в гелии при комнатной температуре, если теплоемкость одноатомного идеального газа
Сv = 3/2 R, атомный вес МНе = 4.

Решение. ( ∂P ∂ρ ) S = ( ∂P ∂( M V ) ) S =− V 2 M ( ∂P ∂V ) S =− V 2 M ∂(P,S) ∂(V,S) . MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@63D4@

∂(P,S) ∂(V,S) = ∂(P,S) ∂(T,P) ∂(T,P) ∂(T,V) ∂(T,V) ∂(V,S) = c P c V ( ∂P ∂V ) T MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6C4B@

c=V ( − 1 M c P c V ( ∂P ∂V ) T ) 0,5 = 1 M c P c V RT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5473@

c= 1 4⋅ 10 −3 5 3 8,314⋅298 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabg2da9maakaaabaWaaSaaaeaacaaIXaaabaGaaGinaiabgwSixlaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIZaaaaaaakmaalaaabaGaaGynaaqaaiaaiodaaaGaaGioaiaacYcacaaIZaGaaGymaiaaisdacqGHflY1caaIYaGaaGyoaiaaiIdaaSqabaaaaa@48DE@ м/с = 1016 м/с

37. (2/1-98).* Вычислить изменение потенциала Гиббса в процессе затвердевания 1 кг переохлажденного бензола при 268,2 К. Давление насыщенного пара твердого бензола при 268,2 К 2279,8 Па, а над жидким бензолом при этой же температуре – 2639,7 Па. Вывести формулы для расчета. Пары бензола считать идеальным газом.

Решение. Задача может быть решена через химические потенциалы, однако в этом разделе предполагается, что студент не знаком еще с этим понятием.

Изменением потенциала Гиббса в процессе Ж → Т может быть представлено как сумма ΔG в последовательных процессах: 1) испарения до достижения равновесия (P1 = Pн.п.ж = 2639,7 Па); 2) изотермическое расширение пара до P3 = Рн.п.т = 2279,8 Па; 3) равновесная кристаллизация насыщенного пара в твердую фазу.

Δ Ж→Т G= Δ 1 G+ Δ 2 G+ Δ 3 G MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadAbbcqGHsgIRcaWGIqaabeaakiaadEeacqGH9aqpcqqHuoardaWgaaWcbaGaaGymaaqabaGccaWGhbGaey4kaSIaeuiLdq0aaSbaaSqaaiaaikdaaeqaaOGaam4raiabgUcaRiabfs5aenaaBaaaleaacaaIZaaabeaakiaadEeaaaa@47C2@ .

Δ 1 G MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaaigdaaeqaaOGaam4raaaa@3910@ и Δ 3 G MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaaiodaaeqaaOGaam4raaaa@3912@ = 0, так как фазовые переходы осуществляются при Р и Т, соответствующих равновесному сосуществованию фаз.

Δ 2 G= ∫ P 1 P 3 ( ∂G ∂P ) T = ∫ P 1 P 3 VdP MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaaikdaaeqaaOGaam4raiabg2da9maapehabaWaaeWaaeaadaWcaaqaaiabgkGi2kaadEeaaeaacqGHciITcaWGqbaaaaGaayjkaiaawMcaamaaBaaaleaacaWGubaabeaaaeaacaWGqbWaaSbaaWqaaiaaigdaaeqaaaWcbaGaamiuamaaBaaameaacaaIZaaabeaaa0Gaey4kIipakiabg2da9maapehabaGaamOvaiaadsgacaWGqbaaleaacaWGqbWaaSbaaWqaaiaaigdaaeqaaaWcbaGaamiuamaaBaaameaacaaIZaaabeaaa0Gaey4kIipaaaa@50A0@ . Для идеального газа PV = RT и Δ 2 G=RTln⁡ P 3 P 1 =8,314⋅268,2⋅ln⁡ 2279,8 2639,7 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaaikdaaeqaaOGaam4raiabg2da9iaadkfacaWGubGaciiBaiaac6gadaWcaaqaaiaadcfadaWgaaWcbaGaaG4maaqabaaakeaacaWGqbWaaSbaaSqaaiaaigdaaeqaaaaakiabg2da9iaaiIdacaGGSaGaaG4maiaaigdacaaI0aGaeyyXICTaaGOmaiaaiAdacaaI4aGaaiilaiaaikdacqGHflY1ciGGSbGaaiOBamaalaaabaGaaGOmaiaaikdacaaI3aGaaGyoaiaacYcacaaI4aaabaGaaGOmaiaaiAdacaaIZaGaaGyoaiaacYcacaaI3aaaaaaa@5904@ Дж/моль = – 326,84 Дж/моль

1 кг бензола – это 12,82 моль и Δ 2 G= MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaaikdaaeqaaOGaam4raiabg2da9aaa@3A17@ 4,19 кДж.

45. (3/1-06).* Оценить величину энергии связи в молекуле О2, если известно, что изобарный тепловой эффект каталитической реакции окисления орто-ксилола до фталевой кислоты, записываемой уравнением

о-С8Н10(ж.) + 6О(г.) = С8Н6О4(кр.) + 2Н2О(ж.),

равен –2824,49 кДж/моль. Теплота сгорания фталевой кислоты равна 3223,33 кДж/моль.

 

Δ f H 298 o MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadAgaaeqaaOGaamisamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaWGVbaaaaaa@3CA3@ , кДж/моль

S 298 o MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaWGVbaaaaaa@3A27@ ,

 C p, 298 o MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDaaaleaacaWGWbGaaiilaiaaykW7caaIYaGaaGyoaiaaiIdaaeaacaWGVbaaaaaa@3D47@ ,

Δ исп H MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadIdbcaWGbrGaam4peaqabaGccaWGibaaaa@3AA9@ , кДж/моль

tкип, оС

Дж/моль×К

CО2 (г)
Н2О (ж)
Н2О (г)

–393,51
–285,83
–241,82

213,79
70,08
188,72

37,14
75,3
33,6

–
40,66
40,66

–
100
100

о–ксилол(ж.)

–24,43

247

187,0

36,24

144

Решение. Сначала определим ΔfHф.к.. По условию, для реакции
С8H6O4(кр.) + 7,5О2 = 8СО2 + 3Н2О ΔrH..= – 3223,.33 кДж/моль.

Δ r H 298 0 = ∑ i ν i Δ f H 298 o MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamisamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaaIWaaaaOGaeyypa0ZaaabuaeaacqaH9oGBdaWgaaWcbaGaamyAaaqabaaabaGaamyAaaqab0GaeyyeIuoakiabfs5aenaaBaaaleaacaWGMbaabeaakiaadIeadaqhaaWcbaGaaGOmaiaaiMdacaaI4aaabaGaam4Baaaaaaa@4A1D@ или – 3223,33 = – 8*393,51 – 3*285,83 – ΔfHф.к., ΔfHф.к.= –782,24 кДж/моль.

Теперь найдем ΔfHО:

о-С8Н10(ж.) + 6О(г.) = С8Н6О4(кр.) + 2Н2О(ж.) ΔrH = –2824,49 кДж/моль

–2824,49= –782,24 – 2*285,83 + 24,43 – 6*ΔfHО,

ΔfHО = 249,17 кДж/моль.

Энергия связи – это энергия диссоциации по реакции O2 = 2 O(г.)

Eсв ≈ Δ rU = ΔrH – ∆r ν.RT = 2*ΔfHО – ∆ rν.RT = 495,86 кДж/моль.


2. Химическое равновесие в гомогенных системах

52. (4/1-97).Для процесса диссоциации идеального газа А2 = 2А выразить в явном виде зависимость константы равновесия KP от степени диссоциации a , измеряемой в изобарном и изохорном процессах. При каком начальном давлении Po(A2) будет достигаться a = 0,5 в этих случаях, если KP = 1 бар?

Решение.

При проведении процесса в изобарных условиях:

KP = P A 2 P A 2 = P Σ (2ξ) 2 ( n 0 −ξ)( n 0 +ξ) = P Σ (2ξ) 2 ( n 0 ) 2 − ξ 2 = P Σ (2α) 2 1− α 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6C4E@ ,

откуда при KP = 1 и α = 0,5, P0(A2) = PS = 0,75 бар.

Если процесс вести изохорно, то в равновесном состоянии

KP = P A 2 P A 2 = RT V (2ξ) 2 ( n 0 −ξ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGqbWaa0baaSqaaiaadgeaaeaacaaIYaaaaaGcbaGaamiuamaaBaaaleaacaWGbbWaaSbaaWqaaiaaikdaaeqaaaWcbeaaaaGccqGH9aqpdaWcaaqaaiaadkfacaWGubaabaGaamOvaaaadaWcaaqaaiaacIcacaaIYaGaeqOVdGNaaiykamaaCaaaleqabaGaaGOmaaaaaOqaaiaacIcacaWGUbWaaWbaaSqabeaacaaIWaaaaOGaeyOeI0IaeqOVdGNaaiykaaaaaaa@49B9@ = n 0 RT V (2α) 2 (1−α) = P A 2 нач (2α) 2 (1−α) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGUbWaaWbaaSqabeaacaaIWaaaaOGaamOuaiaadsfaaeaacaWGwbaaamaalaaabaGaaiikaiaaikdacqaHXoqycaGGPaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaiikaiaaigdacqGHsislcqaHXoqycaGGPaaaaiabg2da9iaadcfadaqhaaWcbaGaamyqamaaBaaameaacaaIYaaabeaaaSqaaiaad2dbcaWGWqGaam4reaaakmaalaaabaGaaiikaiaaikdacqaHXoqycaGGPaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaiikaiaaigdacqGHsislcqaHXoqycaGGPaaaaaaa@5335@ .

После подстановки KP = 1 и α = 0,5 находим: P0(A2) = 0,5 бар.

67. (3/1-99)*. Для реакции диссоциации Br2 (газ) = 2Br (газ)

зависимость константы равновесия от температуры в единицах СИ имеет вид: ln KP = –23009/T + 0,663 lnT + 8,12

Оценить энергию связи в молекуле Br2.

Решение. K P =exp⁡( − Δ r G 0 RT )=exp⁡( − Δ r H 0 (T) RT )exp⁡( Δ r S 0 (T) R ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5EA2@

ln⁡ K P =− Δ r H 0 (T) RT + Δ r S 0 (T) R =− Δ r H 298 0 + Δ r c p ⋅( T−298 ) RT + Δ r S 298 0 + Δ r c p ⋅ln⁡ T 298 R =    =− Δ r H 298 0 −298⋅ Δ r c p RT + Δ r c p R ln⁡T+( Δ r c p ( 1−ln⁡298 )+ Δ r S 298 0 R ). MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@B3E8@

Δ r H 298 0 ≈(23009+298⋅0,663)R=192,94 кДж/моль. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamisamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaaIWaaaaOGaeyisISRaaiikaiaaikdacaaIZaGaaGimaiaaicdacaaI5aGaey4kaSIaaGOmaiaaiMdacaaI4aGaeyyXICTaaGimaiaacYcacaaI2aGaaGOnaiaaiodacaGGPaGaamOuaiabg2da9iaaigdacaaI5aGaaGOmaiaacYcacaaI5aGaaGinaiaaykW7caqG6qGaaeifeiaabAdbcaqGVaGaaeipeiaab6dbcaqG7qGaaeiteiaab6caaaa@5AB9@

Энергию связи в современной справочной литературе определяют как Δ r U 298 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamyvamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaaIWaaaaaaa@3C82@ (для стандартного состояния вещества):

Δ r U 298 0 = Δ r H 298 0 − Δ r ( PV ) 298 = Δ r H 298 0 −298R=190,46  кДж/моль; MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6819@

либо как Δ r U 0K 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamyvamaaDaaaleaacaaIWaGaam4saaqaaiaaicdaaaaaaa@3BCB@ = 23009 R = 191,3 кДж/моль

72. (4/1-08).Газообразный углеводород A участвует в двух реакциях, приводящих к получению изомеров В и С:

A ↔ B и A ↔ С.

Значения стандартных энтальпий, энтропий и потенциалов Гиббса образования указанных веществ при 1000 К приведены в таблице

Вещество

ΔfНo(газ),

ΔfGo(газ),

So(газ),

кДж/моль

Дж/моль·К

А (Перилен)

B (Бензопирен – е)

C (Бензопирен – а)

253,2

253,2

262,4

–734,7

–740,5

–737,0

987,9

993,7

999,4

Определите равновесный состав при 1000 К. Какой из изомеров будет преобладать при последующем повышении температуры?

Решение.

K P1 =exp⁡( − Δ r1 G 0 RT )=exp⁡( − −740500+734700 8,31⋅1000 )=exp⁡(+0,698)=2. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBaaaleaacaWGqbGaaGymaaqabaGccqGH9aqpciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTmaalaaabaGaeuiLdq0aaSbaaSqaaiaadkhacaaIXaaabeaakiaadEeadaahaaWcbeqaaiaaicdaaaaakeaacaWGsbGaamivaaaaaiaawIcacaGLPaaacqGH9aqpciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTmaalaaabaGaeyOeI0IaaG4naiaaisdacaaIWaGaaGynaiaaicdacaaIWaGaey4kaSIaaG4naiaaiodacaaI0aGaaG4naiaaicdacaaIWaaabaGaaGioaiaacYcacaaIZaGaaGymaiabgwSixlaaigdacaaIWaGaaGimaiaaicdaaaaacaGLOaGaayzkaaGaeyypa0JaciyzaiaacIhacaGGWbGaaiikaiabgUcaRiaaicdacaGGSaGaaGOnaiaaiMdacaaI4aGaaiykaiabg2da9iaaikdacaGGUaaaaa@6B14@

K P2 =exp⁡( − Δ r2 G 0 RT )=exp⁡( − −737000+734700 8,31⋅1000 )=exp⁡(+0,277)=1,32. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBaaaleaacaWGqbGaaGOmaaqabaGccqGH9aqpciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTmaalaaabaGaeuiLdq0aaSbaaSqaaiaadkhacaaIYaaabeaakiaadEeadaahaaWcbeqaaiaaicdaaaaakeaacaWGsbGaamivaaaaaiaawIcacaGLPaaacqGH9aqpciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTmaalaaabaGaeyOeI0IaaG4naiaaiodacaaI3aGaaGimaiaaicdacaaIWaGaey4kaSIaaG4naiaaiodacaaI0aGaaG4naiaaicdacaaIWaaabaGaaGioaiaacYcacaaIZaGaaGymaiabgwSixlaaigdacaaIWaGaaGimaiaaicdaaaaacaGLOaGaayzkaaGaeyypa0JaciyzaiaacIhacaGGWbGaaiikaiabgUcaRiaaicdacaGGSaGaaGOmaiaaiEdacaaI3aGaaiykaiabg2da9iaaigdacaGGSaGaaG4maiaaikdacaGGUaaaaa@6D38@

Состав смеси:

x A = P A P A + P B + P C = 1 1+ K P1 + K P2 =0,23 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaWGbbaabeaakiabg2da9maalaaabaGaamiuamaaBaaaleaacaWGbbaabeaaaOqaaiaadcfadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGqbWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaamiuamaaBaaaleaacaWGdbaabeaaaaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaey4kaSIaam4samaaBaaaleaacaWGqbGaaGymaaqabaGccqGHRaWkcaWGlbWaaSbaaSqaaiaadcfacaaIYaaabeaaaaGccqGH9aqpcaaIWaGaaiilaiaaikdacaaIZaaaaa@4F6D@

x B = P B P A + P B + P C = K P1 1+ K P1 + K P2 =0,46 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaWGcbaabeaakiabg2da9maalaaabaGaamiuamaaBaaaleaacaWGcbaabeaaaOqaaiaadcfadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGqbWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaamiuamaaBaaaleaacaWGdbaabeaaaaGccqGH9aqpdaWcaaqaaiaadUeadaWgaaWcbaGaamiuaiaaigdaaeqaaaGcbaGaaGymaiabgUcaRiaadUeadaWgaaWcbaGaamiuaiaaigdaaeqaaOGaey4kaSIaam4samaaBaaaleaacaWGqbGaaGOmaaqabaaaaOGaeyypa0JaaGimaiaacYcacaaI0aGaaGOnaaaa@514F@

x C = P C P A + P B + P C = K P2 1+ K P1 + K P2 =0,3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaWGdbaabeaakiabg2da9maalaaabaGaamiuamaaBaaaleaacaWGdbaabeaaaOqaaiaadcfadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGqbWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaamiuamaaBaaaleaacaWGdbaabeaaaaGccqGH9aqpdaWcaaqaaiaadUeadaWgaaWcbaGaamiuaiaaikdaaeqaaaGcbaGaaGymaiabgUcaRiaadUeadaWgaaWcbaGaamiuaiaaigdaaeqaaOGaey4kaSIaam4samaaBaaaleaacaWGqbGaaGOmaaqabaaaaOGaeyypa0JaaGimaiaacYcacaaIZaaaaa@5091@

При увеличении температуры:

∂ ∂T ( x B x A )= ∂ ∂T ( P B P A )= ∂ K P1 ∂T = K P1 Δ r1 H 0 R T 2 =0. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5E43@

∂ ∂T ( x C x A )= ∂ ∂T ( P C P A )= ∂ K P2 ∂T = K P2 Δ r2 H 0 R T 2 = K P2 ⋅1,1⋅ 10 −3  > 0. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F0F@

В равновесной смеси при повышении температуры будет повышаться мольная доля вещества С, соотношение B:A будет постоянным. Однако рост KP2 очень “медленный” – порядка 0,1 % на 1 K, поэтому преобладать будет изомер B.

96. (4/2-00).Для систем

CuSO4× 5H2O = CuSO4× 3H2O + 2H2O (1)

CuSO4× 3H2O = CuSO4× H2O + 2H2O (2)

CuSO4× H2O = CuSO4 + H2O (3)

давление насыщенного пара при 50 °С равно соответственно 47, 30 и 4,4 торр. В герметичный сосуд небольшого объема поместили 1 моль безводного сульфата меди, откачали и затем стали вводить пары воды при данной температуре. Нарисовать зависимость давления паров воды в системе от количества введенных молей воды. Пояснить приведенный рисунок.

Решение. При давлении паров воды менее 4,4 торр, насыщение парами воды не достигается ни д ля одной реакции. Давление паров воды растет пропорционально количеству введенной воды. После достижения давления 4,4 торр (nH2O = δ) равновесно сосуществуют моногидрат сульфата меди и безводный сульфат, соотношение этих фаз определяется количеством введенной воды, но, пока сосуществуют обе фазы, давление паров воды составляет 4,4 торр. Максимальное количество воды, при вводе которого РH2O = 4,4 торр, составяет δ + 1 моль.

Дальнейшее добавление в систему воды приведет к дальнейшему росту давления паров воды, пока не будет достигнуто давление 30 торр (количество введенной воды составит 6,8δ + 1 моль). При дальнейшем добавлении воды происходит превращение моногидрата в тригидрат. Сосуществование этих фаз в равновесии требует давления паров воды 30 торр. Полное превращение моногидрата в тригидрат будет достигнуто при nH2O = 6,8.δ + 3 моль.

При достижении давления 47 торр в равновесии находится реакция (1). Равновесное давление 47 торр поддерживается в диапазоне введенной воды от 10,7.δ + 3 моль до 10,7.δ + 5 моль. Дальше давление паров воды опять будет линейно повышаться с увеличением количества введенной воды.

107. (2/1-07).Зависимости Δ r G 0 (T) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiaacIcacaWGubGaaiykaaaa@3C6F@ реакций окисления ряда металлов, графита и СО (диаграммы Эллингхэма) приведены на рисунке. Определите: 1) при какой T и какие металлы могут самопроизвольно восста-навливаться из соответ-ствующих оксидов;
2) при какой T и какие металлы можно восста-новить монооксидом углерода;
3) при какой T и какие металлы можно восстано-вить графитом?
Парциальные давления всех газообразных веществ считать равными 1 атм.

Решение. Условие самопроизвольного процесса Δ r G <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaaaaOGaeyipaWJaaGimaaaa@3B41@ , при парциальных давлениях 1 атм это условие переходит в Δ r G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabgYda8iaaicdaaaa@3BFB@ .

Самопроизвольно восстанавливаются:
Ag при Т > 490 К по реакции Ag2O = 2Ag + 0,5O2, − Δ r1 G 0 (T>490)<0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaeuiLdq0aaSbaaSqaaiaadkhacaaIXaaabeaakiaadEeadaqhaaWcbaaabaGaaGimaaaakiaacIcacaWGubGaeyOpa4JaaGinaiaaiMdacaaIWaGaaiykaiabgYda8iaaicdaaaa@4318@ и Cu при Т > 1720 К по реакции CuO = Cu + 0,5O2, − Δ r2 G 0 (T>1720)<0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaeuiLdq0aaSbaaSqaaiaadkhacaaIYaaabeaakiaadEeadaqhaaWcbaaabaGaaGimaaaakiaacIcacaWGubGaeyOpa4JaaGymaiaaiEdacaaIYaGaaGimaiaacMcacqGH8aapcaaIWaaaaa@43D0@ . Cd и Ca не восстанавливаются.

Монооксидом углерода восстанавливаются Ag, Cu и Сd при любой температуре: Ag2O +CO = 2Ag + CO2, Δ r G 0 = Δ r4 G 0 − Δ r1 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGinaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaigdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4807@ ,

CuO + CO = Cu + CO2, Δ r G 0 = Δ r4 G 0 − Δ r2 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGinaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaikdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4808@ ,

Δ r G 0 = Δ r4 G 0 − Δ r3 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGinaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaiodaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4809@ .

Не восстанавливается кальций.

Графитом восстанавливаются:

Ag, при любой T: Ag2O + C = 2Ag + CO, Δ r G 0 = Δ r5 G 0 − Δ r1 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGynaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaigdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4808@ ,

Cu при Т > 280 К: CuO + C = Cu + CO, Δ r G 0 = Δ r5 G 0 − Δ r2 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGynaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaikdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4809@ ,

Cd при Т > 850 К: CdO + C = Cd + CO, Δ r G 0 = Δ r5 G 0 − Δ r3 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGynaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaiodaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@480A@ .

Не восстанавливается кальций.


3. Химическое равновесие в гетерогенных системах

110. (1/2-00). Зависимость температуры плавления Sn от давления (в бар) передается выражением: t(°С) = 231,8 + 0,0032(P–1).
Найти плотность твердого олова r тв, учитывая, что Qпл = 7,2 кДж/моль и r ж = 6,988 г/см3.
Молекулярная масса олова 119.

Решение. Для решения необходимо использовать уравнение Клаузиуса – Клапейрона:   dP dT = Δ ф.п. S ¯ Δ ф.п. V ¯ = Δ ф.п. H ¯ T ф.п. Δ ф.п. V ¯ = Δ пл H ¯ T пл M( 1 ρ ж − 1 ρ тв ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaamiuaaqaaiaadsgacaWGubaaaiabg2da9maalaaabaGaeuiLdq0aaSbaaSqaaiaadsebcaGGUaGaam4peiaac6caaeqaaOWaa0aaaeaacaWGtbaaaaqaaiabfs5aenaaBaaaleaacaWGerGaaiOlaiaad+dbcaGGUaaabeaakmaanaaabaGaamOvaaaaaaGaeyypa0ZaaSaaaeaacqqHuoardaWgaaWcbaGaamireiaac6cacaWG=qGaaiOlaaqabaGcdaqdaaqaaiaadIeaaaaabaGaamivamaaBaaaleaacaWGerGaaiOlaiaad+dbcaGGUaaabeaakiabfs5aenaaBaaaleaacaWGerGaaiOlaiaad+dbcaGGUaaabeaakmaanaaabaGaamOvaaaaaaGaeyypa0ZaaSaaaeaacqqHuoardaWgaaWcbaGaam4peiaadUdbaeqaaOWaa0aaaeaacaWGibaaaaqaaiaadsfadaWgaaWcbaGaam4peiaadUdbaeqaaOGaamytamaabmaabaWaaSaaaeaacaaIXaaabaGaeqyWdi3aaSbaaSqaaiaadAdbaeqaaaaakiabgkHiTmaalaaabaGaaGymaaqaaiabeg8aYnaaBaaaleaacaWGcrGaamOmeaqabaaaaaGccaGLOaGaayzkaaaaaaaa@686A@ .

dP dT = 1 0.0032 бар/К=3,12⋅ 10 7 Па/К; MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaamiuaaqaaiaadsgacaWGubaaaiabg2da9maalaaabaGaaGymaaqaaiaaicdacaGGUaGaaGimaiaaicdacaaIZaGaaGOmaaaacaqGXqGaaeimeiaabcebcaqGVaGaaeOgeiabg2da9iaaiodacaGGSaGaaGymaiaaikdacqGHflY1caaIXaGaaGimamaaCaaaleqabaGaaG4naaaakiaab+bbcaqGWqGaae4laiaabQbbcaGG7aaaaa@4F40@        Δ пл H ¯ =7,2 кДж/моль MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaad+dbcaWG7qaabeaakmaanaaabaGaamisaaaacqGH9aqpcaaI3aGaaiilaiaaikdacaaMe8UaaeOoeiaabsbbcaqG2qGaae4laiaabYdbcaqG+qGaae4oeiaabYebaaa@449B@ ;

Tпл = 505 K; M = 0.119 кг/моль; ρ ж MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadAdbaeqaaaaa@3898@ = 6,988 103 кг/м3

Отсюда следует ответ: ρ тв MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadkebcaWGYqaabeaaaaa@395F@ = 7,18 103 кг/м3

Обратите внимание: в справочнике при 298 К ρ тв MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadkebcaWGYqaabeaaaaa@395F@ = 7,37 103 кг/м3.
Найденная нами величина – это плотность при Тпл = 505 K. Коэффициент теплового расширения (298 К) <αL(Sn) = 2,2.10–5 K–1.
Из этой величины ρ тв MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadkebcaWGYqaabeaaaaa@395F@ (505 K) = ρ тв MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadkebcaWGYqaabeaaaaa@395F@ (298 K).0,986 = 7,26.103 кг/м3.
Причина несовпадения на 40 % по ΔплV – приближенность линейной аппроксимации t(P) в условии задачи

153. (4/2-03). Равновесное давление пара над сконденсированным газом В равно 38 бар при 22 °С. Зависимость коэффициента активности нелетучего растворителя А от состава бинарного жидкого раствора, образованного веществами А и В, при этой температуре описывается уравнением ln⁡ γ A =−0,1 x B 2 −0,8 x B 3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaamyqaaqabaGccqGH9aqpcqGHsislcaaIWaGaaiilaiaaigdacaWG4bWaa0baaSqaaiaadkeaaeaacaaIYaaaaOGaeyOeI0IaaGimaiaacYcacaaI4aGaamiEamaaDaaaleaacaWGcbaabaGaaG4maaaaaaa@470A@ . Определите численное значение константы Генри. Определите состав раствора, находящего в равновесии с газом В, если парциальное давление последнего равно 76 торр.

Решение:

Из выражения для коэффициента активности γ1 компонента 1 бинарного раствора, используя уравнение Гиббса – Дюгема x 1 dln⁡ γ 1 + x 2 dln⁡ γ 2 =0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaaIXaaabeaakiaadsgaciGGSbGaaiOBaiabeo7aNnaaBaaaleaacaaIXaaabeaakiabgUcaRiaadIhadaWgaaWcbaGaaGOmaaqabaGccaWGKbGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaaIWaaaaa@4737@ , можно вывести выражение для γ1.

Выведем это выражение в общем виде для ln⁡ γ 1 = b 1 x 2 2 + c 1 x 2 3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaWGIbWaaSbaaSqaaiaaigdaaeqaaOGaamiEamaaDaaaleaacaaIYaaabaGaaGOmaaaakiabgUcaRiaadogadaWgaaWcbaGaaGymaaqabaGccaWG4bWaa0baaSqaaiaaikdaaeaacaaIZaaaaaaa@4551@ .

x 1 dln⁡ γ 1 + x 2 dln⁡ γ 2 = x 1 (2 b 1 x 2 +3 c 1 x 2 )d x 2 + x 2 dln⁡ γ 2 =0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6172@

dln⁡ γ 2 =−(2 b 1 x 1 +3 c 1 x 1 x 2 )d x 2 =(2 b 1 x 1 +3 c 1 x 1 (1− x 1 ))d x 1 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6151@

ln⁡ γ 2 = ∫ (−3 c 1 x 1 2 +(2 b 1 +3 c 1 ) x 1 )d x 1 = ( b 1 + 3 2 c 1 ) x 1 2 −3 c 1 x 1 3 +C MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6507@

При x 1 →0,  γ 2 →1,ln⁡ γ 2 →0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaaIXaaabeaakiabgkziUkaaicdacaGGSaGaaGPaVlabeo7aNnaaBaaaleaacaaIYaaabeaakiabgkziUkaaigdacaGGSaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGOmaaqabaGccqGHsgIRcaaIWaaaaa@49D2@ и, следовательно, С = 0.

ln⁡ γ 2 = b 2 x 1 2 + c 2 x 1 3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaWGIbWaaSbaaSqaaiaaikdaaeqaaOGaamiEamaaDaaaleaacaaIXaaabaGaaGOmaaaakiabgUcaRiaadogadaWgaaWcbaGaaGOmaaqabaGccaWG4bWaa0baaSqaaiaaigdaaeaacaaIZaaaaaaa@4552@ , b 2 = b 1 + 3 2 c 1 , c 2 =− c 1 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOyamaaBaaaleaacaaIYaaabeaakiabg2da9iaadkgadaWgaaWcbaGaaGymaaqabaGccqGHRaWkdaWcaaqaaiaaiodaaeaacaaIYaaaaiaadogadaWgaaWcbaGaaGymaaqabaGccaGGSaGaam4yamaaBaaaleaacaaIYaaabeaakiabg2da9iabgkHiTiaadogadaWgaaWcbaGaaGymaaqabaaaaa@4534@ .

Для условия данной задачи ln⁡ γ B =−0,7 x A 2 +0,8 x A 3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaamOqaaqabaGccqGH9aqpcqGHsislcaaIWaGaaiilaiaaiEdacaWG4bWaa0baaSqaaiaadgeaaeaacaaIYaaaaOGaey4kaSIaaGimaiaacYcacaaI4aGaamiEamaaDaaaleaacaWGbbaabaGaaG4maaaaaaa@4704@ .

Константа Генри – это значение в предельно разбавленном растворе
K Г = lim⁡ x B →0 ( P B 0 γ B )= lim⁡ x A →1 ( P B 0 γ B )=exp⁡(0,1) P B 0 =42  атм. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@68F2@

161. (3/2-05).* Найти работу разделения эквимолярной бинарной неидеальной смеси на чистые компоненты при 298 К и атмосферном давлении, если известно, что логарифм коэффициента активности γ1 для этой смеси зависит от состава согласно уравнению:

ln⁡ γ 1 = b 1 x 2 2 + c 1 x 2 3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaWGIbWaaSbaaSqaaiaaigdaaeqaaOGaamiEamaaDaaaleaacaaIYaaabaGaaGOmaaaakiabgUcaRiaadogadaWgaaWcbaGaaGymaaqabaGccaWG4bWaa0baaSqaaiaaikdaaeaacaaIZaaaaaaa@4551@ ,  где b 1 =0,5, c 1 =1. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOyamaaBaaaleaacaaIXaaabeaakiabg2da9iaaicdacaGGSaGaaGynaiaacYcacaWGJbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0JaaGymaiaac6caaaa@3FF0@

Решение. Из выражения для коэффициента активности γ1 компонента 1 бинарного раствора, используя уравнение Гиббса – Дюгема x 1 dln⁡ γ 1 + x 2 dln⁡ γ 2 =0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaaIXaaabeaakiaadsgaciGGSbGaaiOBaiabeo7aNnaaBaaaleaacaaIXaaabeaakiabgUcaRiaadIhadaWgaaWcbaGaaGOmaaqabaGccaWGKbGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaaIWaaaaa@4737@ ,

можно вывести выражение для γ2 (см. выше решение задачи 155).

ln⁡ γ 2 = b 2 x 1 2 + c 2 x 1 3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacqaHZoWzdaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaWGIbWaaSbaaSqaaiaaikdaaeqaaOGaamiEamaaDaaaleaacaaIXaaabaGaaGOmaaaakiabgUcaRiaadogadaWgaaWcbaGaaGOmaaqabaGccaWG4bWaa0baaSqaaiaaigdaaeaacaaIZaaaaaaa@4552@ , b 2 = b 1 + 3 2 c 1 , c 2 =− c 1 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOyamaaBaaaleaacaaIYaaabeaakiabg2da9iaadkgadaWgaaWcbaGaaGymaaqabaGccqGHRaWkdaWcaaqaaiaaiodaaeaacaaIYaaaaiaadogadaWgaaWcbaGaaGymaaqabaGccaGGSaGaam4yamaaBaaaleaacaaIYaaabeaakiabg2da9iabgkHiTiaadogadaWgaaWcbaGaaGymaaqabaaaaa@4534@ .

Для условия этой задачи b 2 =2, c 2 =−1 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOyamaaBaaaleaacaaIYaaabeaakiabg2da9iaaikdacaGGSaGaam4yamaaBaaaleaacaaIYaaabeaakiabg2da9iabgkHiTiaaigdaaaa@3EC0@ . Для эквимолярной смеси

ln⁡ γ 1 =0,5 x 2 2 + x 2 3 =0,25,  ln⁡ γ 2 =2 x 1 2 − x 1 3 =0,375 γ 1 =1,28,  γ 2 =1,455. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6D7E@

A=− G M =−RT( x 1 ln⁡ x 1 γ 1 + x 2 ln⁡ x 2 γ 2 )=0,068RT=168,4 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2da9iabgkHiTiaadEeadaahaaWcbeqaaiaad2eaaaGccqGH9aqpcqGHsislcaWGsbGaamivaiaacIcacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaciiBaiaac6gacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaeq4SdC2aaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiEamaaBaaaleaacaaIYaaabeaakiGacYgacaGGUbGaamiEamaaBaaaleaacaaIYaaabeaakiabeo7aNnaaBaaaleaacaaIYaaabeaakiaacMcacqGH9aqpcaaIWaGaaiilaiaaicdacaaI2aGaaGioaiaadkfacaWGubGaeyypa0JaaGymaiaaiAdacaaI4aGaaiilaiaaisdaaaa@5C19@ Дж/моль

180. (4/2-05).* На рисунке приведена диаграмма состояния “температура – состав” для системы вода-фенол при давлении 760 торр. Определите, какие и какого состава фазы присутствуют в системе в точках А – Е. Нарисуйте схематично диаграмму при P = 100 торр.
Для воды D испН = 40,66 кДж/моль (tкип = 100 оС).
Для фенола D испН = 47,97 кДж/моль (tкип = 181,9 оС).

Решение:

А – пар ~ 70 % Н2О.

В – пар ~ 70 % Н2О и раствор воды в феноле (x(Н2О) ~ 5 %).

С – раствор (x(Н2О) ~ 50 %)

D = С

E – раствор фенола в воде (x(H2O) ~ 96 %) и раствор воды в феноле (x(Н2О) ~ 73 %).

При давлении 100 торр температуры кипения чистых компонентов можно оценить по формуле

ln⁡( P 2 P 1 )=− Δ исп H R ( 1 T 2 − 1 T 1 ), MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gadaqadaqaamaalaaabaGaamiuamaaBaaaleaacaaIYaaabeaaaOqaaiaadcfadaWgaaWcbaGaaGymaaqabaaaaaGccaGLOaGaayzkaaGaeyypa0JaeyOeI0YaaSaaaeaacqqHuoardaWgaaWcbaGaamioeiaadgebcaWG=qaabeaakiaadIeaaeaacaWGsbaaamaabmaabaWaaSaaaeaacaaIXaaabaGaamivamaaBaaaleaacaaIYaaabeaaaaGccqGHsisldaWcaaqaaiaaigdaaeaacaWGubWaaSbaaSqaaiaaigdaaeqaaaaaaOGaayjkaiaawMcaaiaacYcaaaa@4CDE@   которая получается интегрированием уравнения Клапейрона – Клаузиуса для фазового перехода “жидкость–пар” в приближении Δ исп H MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadIdbcaWGbrGaam4peaqabaGccaWGibaaaa@3AA9@ = const.

Подставляя данные из задачи, получаем

Tкип(H2O, 100 торр) = 323 K, Tкип(фенол, 100 торр) = 392,2 K

Зависимостью растворимости жидкостей друг в друге от давления можно пренебречь.

Диаграмма при P = 100 торр приведена на рисунке:

 

 

 

 

203. (5/2-07). На рисунке представлена фазовая диаграмма T(x) для бинарной системы “Ar – N2” в области фазовых переходов твердое – жидкость
(T указана в К).
Опишите фазовый состав в точках А–F.

(по данным: Long H. M., Di Paolo F. S., 1963, Chem. Eng. Prog. Symp. Ser., Vol. 59, p. 30–35).

Решение. A – жидкий раствор аргона в азоте (х(N2) = 75 %)

В – твердый раствор Т1 аргона в азоте (х(N2) = 75 %)

С – 2 фазы: жидкость с х(N2) ~ 70 % и твердое Т1 с х(N2) ~ 62 %

соотношение Ж/Т1 ~ (67–62)/(70–67) = 5/3

D – 2 фазы: жидкость с х(N2) ~ 63 % и твердое Т2 с х(N2) ~ 40 %

соотношение Ж/Т2 ~ (55–40)/(63–55) = 15/8

E – 2 фазы: твердое Т1 с х(N2) ~ 61 % и твердое Т2 с х(N2) ~ 54 %

соотношение Т1/Т2 ~ (55–54)/(61–55) = 1/6

F – 1 фаза: твердый раствор Т2 азота в аргоне (х(N2) = 45 %)

При анализе диаграммы необходимо учесть, что в соответствии с правилом фаз Гиббса линия солидуса – горизонтальная прямая, если в граничащих областях диаграммы существуют 3 фазы (например, в области D – Ж и Т2, а в области С – Ж и Т1). Если же линия солидуса – кривая, то в граничащих областях существуют только 2 фазы (например, по этой причине область B может быть только однофазной и содержит твердый раствор Т1).

214. (2/2-07).* Для выделения чистой воды из морской вблизи берега моря вырыт колодец, стенки которого выполнены из материала, который можно рассматривать как идеальную селективную по воде мембрану, проницаемую для воды и не пропускающую растворенные в морской воде соли. Рассчитайте КПД такого способа. Мольную долю ионов в морской воде можно полагать равной 2 %.

Решение. Уровень воды в колодце ниже, чем в море из-за осмотического давления π. Работа, которую необходимо совершить для выделения 1 моля чистой воды, – это работа на подьём воды из колодца на высоту h= π ρg MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiabg2da9maaliaabaGaeqiWdahabaGaeqyWdiNaam4zaaaaaaa@3C5B@  против силы тяжести F=gρ V ¯ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2da9iaadEgacqaHbpGCdaqdaaqaaiaadAfaaaaaaa@3B56@ . Если КПД насоса η, то совершенная работа составит
W=ηπ V ¯ =η( CRT ) V ¯ =η( x V ¯ RT ) V ¯ =ηxRT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vaiabg2da9iabeE7aOjabec8aWnaanaaabaGaamOvaaaacqGH9aqpcqaH3oaAdaqadaqaaiaadoeacaWGsbGaamivaaGaayjkaiaawMcaamaanaaabaGaamOvaaaacqGH9aqpcqaH3oaAdaqadaqaamaalaaabaGaamiEaaqaamaanaaabaGaamOvaaaaaaGaamOuaiaadsfaaiaawIcacaGLPaaadaqdaaqaaiaadAfaaaGaeyypa0Jaeq4TdGMaamiEaiaadkfacaWGubaaaa@51F2@

(необходимо, конечно, помнить, что это уравнение для осмоса работает при мольной доле π= RTln⁡(1−x) V ¯ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWdaNaeyypa0ZaaSaaaeaacaWGsbGaamivaiGacYgacaGGUbGaaiikaiaaigdacqGHsislcaWG4bGaaiykaaqaamaanaaabaGaamOvaaaaaaaaaa@413E@ ).

Минимальная требуемая работа для выделения пресной воды из морской:

W min⁡ = G H2O M =RTln⁡(1−x) → x<<1 xRT MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4vamaaBaaaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0Jaam4ramaaDaaaleaacaWGibGaaGOmaiaad+eaaeaacaWGnbaaaOGaeyypa0JaamOuaiaadsfaciGGSbGaaiOBaiaacIcacaaIXaGaeyOeI0IaamiEaiaacMcadaGdOaWcbeqaaiaadIhacqGH8aapcqGH8aapcaaIXaaakiaawkziaiaadIhacaWGsbGaamivaaaa@4FAA@

Следовательно, КПД предлагаемого способа – это КПД насоса, который качает воду из колодца.


4. Химическое равновесие в системах, содержащих растворы электролитов

217. (3/3-94).* В воде растворено некоторое количество нелетучего слабого электролита, не диссоциирующего при низкой температуре, но полностью диссоциирующего на два иона при температурах, близких к температуре кипения воды. Найти температуру кипения данного раствора, если известно, что этот же раствор замерзает при 271,5 К. Учесть, что для чистой воды D Нпл = 6,029 кДж/моль, Тпл = 273,15 К, D Нисп = 40,62 кДж/моль, Ткип = 373,15 К. Определить давление паров воды над раствором при 299 К, если над чистой водой при 298 К оно равно 0,03168 бар.

Решение. Так как раствор замерзает при 271,5 K, то есть на 1,65 K ниже нормальной точки плавления льда, то можно оценить содержание примеси в растворе, предполагая, что во льду она не растворяется, по уравнению Шредера:

ln⁡(1−x)= Δ пл H R ( 1 T пл 0 − 1 T пл ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaGGOaGaaGymaiabgkHiTiaadIhacaGGPaGaeyypa0ZaaSaaaeaacqqHuoardaWgaaWcbaGaam4peiaadUdbaeqaaOGaamisaaqaaiaadkfaaaWaaeWaaeaadaWcaaqaaiaaigdaaeaacaWGubWaa0baaSqaaiaad+dbcaWG7qaabaGaaGimaaaaaaGccqGHsisldaWcaaqaaiaaigdaaeaacaWGubWaaSbaaSqaaiaad+dbcaWG7qaabeaaaaaakiaawIcacaGLPaaaaaa@4BAE@ , мольная доля примеси x = 1,61 %.

При температуре кипения x2 = 2x = 3,22 % и, предполагая раствор предельно разбавленным и γH2O = 1, можно оценить температуру кипения из уравнения

ln⁡(1− x 2 )= Δ исп H R ( 1 T кип − 1 T кип 0 ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaGGOaGaaGymaiabgkHiTiaadIhadaWgaaWcbaGaaGOmaaqabaGccaGGPaGaeyypa0ZaaSaaaeaacqqHuoardaWgaaWcbaGaamioeiaadgebcaWG=qaabeaakiaadIeaaeaacaWGsbaaamaabmaabaWaaSaaaeaacaaIXaaabaGaamivamaaBaaaleaacaWG6qGaamioeiaad+dbaeqaaaaakiabgkHiTmaalaaabaGaaGymaaqaaiaadsfadaqhaaWcbaGaamOoeiaadIdbcaWG=qaabaGaaGimaaaaaaaakiaawIcacaGLPaaaaaa@4EE7@ .

Отсюда вычисляем T кип =374,1 K MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivamaaBaaaleaacaWG6qGaamioeiaad+dbaeqaaOGaeyypa0JaaG4maiaaiEdacaaI0aGaaiilaiaaigdacaaMe8Uaam4saaaa@4052@ .

Давление паров над раствором находим из закона Рауля для растворителя в предельно разбавленных растворах

P н.п. =(1−x) P н.п. 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBaaaleaacaWG9qGaaiOlaiaad+dbcaGGUaaabeaakiabg2da9iaacIcacaaIXaGaeyOeI0IaamiEaiaacMcacaWGqbWaa0baaSqaaiaad2dbcaGGUaGaam4peiaac6caaeaacaaIWaaaaaaa@439C@ , где P н.п. 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaDaaaleaacaWG9qGaaiOlaiaad+dbcaGGUaaabaGaaGimaaaaaaa@3A9B@ определяется из уравнения Клапейрона –Клаузиуса для равновесия "жидкость – идеальный газ" dln⁡ P н.п. 0 dT = Δ исп H R T 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaciiBaiaac6gacaWGqbWaa0baaSqaaiaad2dbcaGGUaGaam4peiaac6caaeaacaaIWaaaaaGcbaGaamizaiaadsfaaaGaeyypa0ZaaSaaaeaacqqHuoardaWgaaWcbaGaamioeiaadgebcaWG=qaabeaakiaadIeaaeaacaWGsbGaamivamaaCaaaleqabaGaaGOmaaaaaaaaaa@47AF@ или P н.п. 0 ( T 2 ) P н.п. 0 ( T 1 ) =exp⁡( Δ исп H R ( T 2 − T 1 T 1 T 2 ) ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGqbWaa0baaSqaaiaad2dbcaGGUaGaam4peiaac6caaeaacaaIWaaaaOGaaiikaiaadsfadaWgaaWcbaGaaGOmaaqabaGccaGGPaaabaGaamiuamaaDaaaleaacaWG9qGaaiOlaiaad+dbcaGGUaaabaGaaGimaaaakiaacIcacaWGubWaaSbaaSqaaiaaigdaaeqaaOGaaiykaaaacqGH9aqpciGGLbGaaiiEaiaacchadaqadaqaamaalaaabaGaeuiLdq0aaSbaaSqaaiaadIdbcaWGbrGaam4peaqabaGccaWGibaabaGaamOuaaaadaqadaqaamaalaaabaGaamivamaaBaaaleaacaaIYaaabeaakiabgkHiTiaadsfadaWgaaWcbaGaaGymaaqabaaakeaacaWGubWaaSbaaSqaaiaaigdaaeqaaOGaamivamaaBaaaleaacaaIYaaabeaaaaaakiaawIcacaGLPaaaaiaawIcacaGLPaaaaaa@5A71@ .

Подставляя данные условия задачи, находим P н.п. 0 (299 K)=1,056⋅ P н.п. 0 (298 K)=0,03347  бар MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaDaaaleaacaWG9qGaaiOlaiaad+dbcaGGUaaabaGaaGimaaaakiaacIcacaaIYaGaaGyoaiaaiMdacaaMc8Uaam4saiaacMcacqGH9aqpcaaIXaGaaiilaiaaicdacaaI1aGaaGOnaiabgwSixlaadcfadaqhaaWcbaGaamypeiaac6cacaWG=qGaaiOlaaqaaiaaicdaaaGccaGGOaGaaGOmaiaaiMdacaaI4aGaaGPaVlaadUeacaGGPaGaeyypa0JaaGimaiaacYcacaaIWaGaaG4maiaaiodacaaI0aGaaG4naiaaykW7caaMc8UaaeymeiaabcdbcaqGaraaaa@5DAB@ .

А давление насыщенных паров над раствором P н.п. =0,03293 бар MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBaaaleaacaWG9qGaaiOlaiaad+dbcaGGUaaabeaakiabg2da9iaaicdacaGGSaGaaGimaiaaiodacaaIYaGaaGyoaiaaiodacaaMc8UaaeymeiaabcdbcaqGaraaaa@43CE@ .

235. (2/3-99).* Цинковый электрод гальванического элемента Якоби находился исходно в 10–3 М растворе ионов Zn2+. Найти, как изменится ЭДС этой ячейки при 25 °С, если к раствору Zn2+ добавили такой же объем 10–3 М раствора ионов Pb2+. Можно полагать, что внутри рассматриваемой части гальванического элемента термодинамическое равновесие устанавливается очень быстро. Известно, что при 298 К E P b 2+ / Pb o MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyramaaDaaaleaadaWcgaqaaiaadcfacaWGIbWaaWbaaWqabeaacaaIYaGaey4kaScaaaWcbaGaamiuaiaadkgaaaaabaGaam4Baaaaaaa@3D3D@ = –0,126 В отн. Н.В.Э., E Z n 2+ / Zn o MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyramaaDaaaleaadaWcgaqaaiaadQfacaWGUbWaaWbaaWqabeaacaaIYaGaey4kaScaaaWcbaGaamOwaiaad6gaaaaabaGaam4Baaaaaaa@3D69@ = –0,763 В отн. Н.В.Э.

Решение. После добавления Pb2+ протекает реакция Pb2+ + Zn0 ↔ Pb0 + Zn2+.

Константа равновесия этой реакции составляет

a Z n 2+ a P b 2+ =exp⁡( nΔ E 0 F RT )=exp⁡( 2(−0,126+0,763)96485 8,314⋅298 )=3,5⋅ 10 21 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@70AB@

и весь свинец из раствора восстановится до металлического состояния, а эквивалентное количество цинка растворится. Окончательная концентрация катионов цинка 10–3 М – такая же, как и была. Следовательно, ЭДС не изменится.

244. (3/3-06).* Произведение растворимости Сu(OH)2 в водном растворе при 25 оС составляет 2.10–20 М3. Оцените, как будет изменяться электродный потенциал электрода Сu2+/Cu при повышении рН, если исходная активность Cu2+ в растворе при рН = 3 составляла 1,0 М. Стандартный электродный потенциал для полуреакции
Сu2+ + 2e– → Cu0 cоставляет + 0,399 В относительно Н.В.Э.

Решение: При активности Сu2+ = 1 M, равновесие Cu(OH)2 ↔ Cu2+ + 2(OH–) достигается при aOH = 1,41.10–10 M.

Поэтому при рН ниже (14 – 9,85 = 4,15) электрод представляет собой электрод первого рода и

Е (pH < 4,15) = + 0,399 В отн. Н.В.Э.

При рН выше 4,15 активностькатионов меди a C u 2+ = 2⋅ 10 −20 a O H − 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaDaaaleaacaWGdbGaamyDamaaCaaameqabaGaaGOmaiabgUcaRaaaaSqaaaaakiabg2da9maalaaabaGaaGOmaiabgwSixlaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaGaaGimaaaaaOqaaiaadggadaqhaaWcbaGaam4taiaadIeadaahaaadbeqaaiabgkHiTaaaaSqaaiaaikdaaaaaaaaa@4764@      ln⁡ a C u 2+ =ln⁡(2⋅ 10 −20 )−2ln⁡ a O H − MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaWGHbWaa0baaSqaaiaadoeacaWG1bWaaWbaaWqabeaacaaIYaGaey4kaScaaaWcbaaaaOGaeyypa0JaciiBaiaac6gacaGGOaGaaGOmaiabgwSixlaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaIYaGaaGimaaaakiaacMcacqGHsislcaaIYaGaciiBaiaac6gacaWGHbWaa0baaSqaaiaad+eacaWGibWaaWbaaWqabeaacqGHsislaaaaleaaaaaaaa@4F46@ = 19,11-4,6 рН

В этом интервале рН электрод представляет собой электрод II рода и

E (pH > 4,15) = E0Cu2+/Cu + RT/2F. (19,1 – 4,6 рН) = (0,644 – 0,059 рН) В отн. Н.В.Э.


5. Термодинамика дисперсных систем и поверхностных явлений

253. (3/3-08).* Суспензия серы в воде была исследована методом термического анализа. Суспензию нагревали до температуры 200 оС в ячейках-автоклавах со скоростью нагрева около 1 К/мин, при этом растворения серы в воде не наблюдалось. Используя приведенную на рисунке кривую ДТА, оцените минимальный размер частиц серы. Коэффициент поверхностного натяжения для серы
σ = 0,042 Дж/м2, ρS = 2,07 г/см3.

Фазовый переход

ΔфпH0.,кДж/моль

Тфп, 0С

S (ромб) ® S (монокл)

0,36

95,6

S (монокл) ® S (ж)

1,26

119,3

H2O (ж) ® H2O (п)

40,4

100

Решение. Первый пик – переход дисперсной ромбической в дисперсную моноклинную, второй пик – плавление серы. Температуру фазового перехода необходимо определять по началу пика. Начало второго пика ~ 115 оС т.е. разница между плавлением дисперсной и массивной серы около 4,3 K.

Используем соотношение для плавления дисперсной фазы:

ln⁡( T T 0 )=− 2σ V ¯ тв r Δ пл H MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gadaqadaqaamaalaaabaGaamivaaqaaiaadsfadaWgaaWcbaGaaGimaaqabaaaaaGccaGLOaGaayzkaaGaeyypa0JaeyOeI0YaaSaaaeaacaaIYaGaeq4Wdm3aa0aaaeaacaWGwbaaamaaBaaaleaacaWGcrGaamOmeaqabaaakeaacaWGYbGaeuiLdq0aaSbaaSqaaiaad+dbcaWG7qaabeaakiaadIeaaaaaaa@4822@ , получаем: r = 10–7 м.

263. (7/П-05). Природа активного компонента Сu-Zn катализатора синтеза метанола вызывает много дискуссий. Одной из гипотез является образование Сu-Zn бронзы при обработке в восстановительной среде. Оценить равновесный состав поверхности Cu-Zn сплава с мольным содержанием Zn 3.10–4 % при температуре 500 К, если известно, что значения избыточной поверхностной энергии для металлических меди и цинка равны соответственно: σСu = 1,3 Дж/м2, а σZn = 0,67 Дж/м2. Сплав меди и цинка полагать идеальным. Концентрация атомов металлов на поверхности сплава равна 1,5.1019 ат..м–2.

Решение. Процесс миграции 1 моля Zn из объема на поверхность бесконечно большого количества частиц Сu-Zn сплава сопровождается изменением потенциала Гиббса:

ΔG=− G M + G M + A ¯ ( σ Zn − σ Cu )= MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam4raiabg2da9iabgkHiTiaadEeadaahaaWcbeqaaiaad2eaaaGcdaqhaaWcbaGaamOwaiaad6gaaeaacaWG+qGaamymeiaadQebcaWG1qGaamipeaaakiabgUcaRiaadEeadaahaaWcbeqaaiaad2eaaaGcdaqhaaWcbaGaamOwaiaad6gaaeaacaWG=qGaamOpeiaadkdbcaWG1qGaamiqeiaadwebaaGccqGHRaWkdaqdaaqaaiaadgeaaaGaaiikaiabeo8aZnaaBaaaleaacaWGAbGaamOBaaqabaGccqGHsislcqaHdpWCdaWgaaWcbaGaam4qaiaadwhaaeqaaOGaaiykaiabg2da9aaa@57B1@

=RT(−ln⁡ x Zn объем +ln⁡ x Zn поверх ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaamOuaiaadsfacaGGOaGaeyOeI0IaciiBaiaac6gacaWG4bWaa0baaSqaaiaadQfacaWGUbaabaGaamOpeiaadgdbcaWGkrGaamyneiaadYdbaaGccqGHRaWkciGGSbGaaiOBaiaadIhadaqhaaWcbaGaamOwaiaad6gaaeaacaWG=qGaamOpeiaadkdbcaWG1qGaamiqeiaadwebaaGccaGGPaaaaa@4E15@ – 0,63.4.104 Дж/моль.

Условие равновесия: ΔG=0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam4raiabg2da9iaaicdaaaa@39DF@ . Решая уравнение, получаем:

x Zn поверх =429⋅ x Zn объем =0,13 % MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaDaaaleaacaWGAbGaamOBaaqaaiaad+dbcaWG+qGaamOmeiaadwdbcaWGarGaamyreaaakiabg2da9iaaisdacaaIYaGaaGyoaiabgwSixlaadIhadaqhaaWcbaGaamOwaiaad6gaaeaacaWG+qGaamymeiaadQebcaWG1qGaamipeaaakiabg2da9iaaicdacaGGSaGaaGymaiaaiodacaaMe8Uaaiyjaaaa@501A@ .

268. (4/3-07).* Железную деталь покрывают краской, состоящей из порошка кадмия. Возможно ли приготовить краску из такого ультрадисперсного порошка кадмия, при использовании которого не будет происходить коррозионного разрушения детали во влажной среде в присутствии кислорода? E0(Cd2+/Cd) = – 0,403 В, E0(Fe2+/Fe) = – 0,447 В отн. Н.В.Э., ρCd = 8,65 г/см3, МCd = 112,41 г/моль, σCd ≈ 0,8 Дж/м2.

Примечание. Коррозия происходит из-за протекания двух полуреакций – анодной Me à Men++ ne– и катодной
O2 + 4H+ + 4e– → 2H2O (E0 = 1,23 В отн. НВЭ).

Решение. Для обеспечения антикоррозионной защиты должно выполняться условие, чтобы электродный потенциал полуреакции восстановления защищаемого металла E0(Fe2+/Fe) был больше потенциала защиты E0(Cd2+/Cd):

Δ E 0 = E Fe 0 − E Cd(дисп.) 0 =0= E Fe 0 − E Cd 0 + 2σ V Cd ¯ 2Fr r= σ M Cd F ρ Cd ( E Cd 0 − E Fe 0 ) =2.7 нм MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@74EA@

270. (5/3-06).* На графике показаны начальные участки изотерм адсорбции 1,3-бутадиена на угле при двух разных температурах. Найти энтальпию адсорбции 1,3-бутадиена на угле, если известно, что она не зависит от степени покрытия поверхности адсорбатом.

Решение:

При одинаковом заполнении поверхности ( dln⁡P dT ) θ = Δ ads H R T 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaadaWcaaqaaiaadsgaciGGSbGaaiOBaiaadcfaaeaacaWGKbGaamivaaaaaiaawIcacaGLPaaadaWgaaWcbaGaeqiUdehabeaakiabg2da9maalaaabaGaeuiLdq0aaSbaaSqaaiaadggacaWGKbGaam4CaaqabaGccaWGibaabaGaamOuaiaadsfadaahaaWcbeqaaiaaikdaaaaaaaaa@47B5@ . Например, при θ = 0,21

ln⁡ 617 300 =− Δ ads H R ( 1 288 − 1 318 ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gadaWcaaqaaiaaiAdacaaIXaGaaG4naaqaaiaaiodacaaIWaGaaGimaaaacqGH9aqpcqGHsisldaWcaaqaaiabfs5aenaaBaaaleaacaWGHbGaamizaiaadohaaeqaaOGaamisaaqaaiaadkfaaaGaaiikamaalaaabaGaaGymaaqaaiaaikdacaaI4aGaaGioaaaacqGHsisldaWcaaqaaiaaigdaaeaacaaIZaGaaGymaiaaiIdaaaGaaiykaaaa@4CAE@

значит, Δ ads H MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadggacaWGKbGaam4CaaqabaGccaWGibaaaa@3B1D@ = –18,3 кДж/моль.


6. Использование статистических методов для описания химического равновесия в идеальных системах

293. (1/3-97).* Используя методы статистической термодинамики, найти температурную зависимость константы равновесия Кр для газофазной диссоциации двухатомной молекулы. Привести график ожидаемого изменения Кр в широкой области температур.

Решение.

A2à 2A

K P =exp⁡( − ΔE RT ) ( kT P 0 V ) Δν ∏ i ( z i ' ) ν i = = kT P 0 V ⋅ z trA 2 z tr A 2 z rot −1 z vib −1 ( 2 s A +1 ) 2 2 s A 2 +1 exp⁡( − ΔE RT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@890C@

z trA 2 z tr A 2 = ( 2πmkT h 2 ) 3 V 2 ( 4πmkT h 2 ) 3/2 V = ( πmkT h 2 ) 3/2 V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@642B@ .

Для вращательной статсуммы при T < θrot
z rot = 1 σ ∑ J−0 ∞ ( 2J+1 ) exp⁡(− h 2 J(J+1) 8 π 2 IkT )= 1 2 ∑ J−0 ∞ ( 2J+1 ) exp⁡(− h 2 J(J+1) 4 π 2 m r 2 kT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7B55@ .

При T >> θrot     z rot = 2 π 2 m r 2 kT h 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWGYbGaam4BaiaadshaaeqaaOGaeyypa0ZaaSaaaeaacaaIYaGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGaamyBaiaadkhadaahaaWcbeqaaiaaikdaaaGccaWGRbGaamivaaqaaiaadIgadaahaaWcbeqaaiaaikdaaaaaaaaa@4503@ .

Для колебательной статсуммы

при T < θvib= hν k MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGObGaeqyVd4gabaGaam4Aaaaaaaa@3992@       z vib = ∑ V−0 ∞ exp⁡( −V hν kT ) = 1 1−exp⁡( − hν kT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaabCaeaaciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTiaadAfadaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaSqaaiaadAfacqGHsislcaaIWaaabaGaeyOhIukaniabggHiLdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaeyOeI0IaciyzaiaacIhacaGGWbWaaeWaaeaacqGHsisldaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaaaaaa@5929@

При T >> θvib      z vib = T θ vib = kT hν MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaSaaaeaacaWGubaabaGaeqiUde3aaSbaaSqaaiaadAhacaWGPbGaamOyaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGRbGaamivaaqaaiaadIgacqaH9oGBaaaaaa@4621@ .

Группируя всё выше, получаем:

при T < θrot

K P = 2 ( πm ) 3 2 h 3 ( 2 s A +1 ) 2 ( 2 s A 2 +1 ) exp⁡( − ΔE RT )⋅ (kT) 5 2 P 0 ⋅( 1−exp⁡( − θ vib T ) ) ∑ J−0 ∞ { ( 2J+1 )exp⁡( − θ rot J(J+1) T ) } ≈const⋅exp⁡( − ΔE RT )⋅ (kT) 5 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@A7B5@

или ln⁡​ K P ≈ − ΔE RT + 2,5 ln⁡Τ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGOmaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C9@

При θrot << T < θvib

K P = m π 1 r 2 h ( 2 s A +1 ) 2 ( 2 s A 2 +1 ) ⋅exp⁡( − ΔE RT )⋅ (kT) 3 2 P 0 ⋅( 1−exp⁡( − θ vib T ) )≈const⋅exp⁡( − ΔE RT ) (kT) 3 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@8904@

или         ln⁡​ K P ≈ − ΔE RT + 1,5 ln⁡Τ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGymaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C8@

При T >> θvib

K P = m π ν vib r 2 ( 2 s A +1 ) 2 ( 2 s A 2 +1 ) ⋅exp⁡( − ΔE RT )⋅ (kT) 1 2 P 0 ⋅=const⋅exp⁡( − ΔE RT ) (kT) 1 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7D0D@

или      ln⁡​ K P ≈ − ΔE RT + 0,5 ln⁡Τ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGimaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C7@

(сравните с задачей 67. (3/1-99), решенной выше:
для диссоциации Br2 ln KP = –23009/T + 0,663 lnT + 8,12)

311. (5/Э-04).* Природное содержание изотопа 13С составляет 1,12 % всего углерода, а дейтерия 2Н – 1,6 10–4 всего водорода. Определить минимальную работу, необходимую для выделения 1 моля “сверхтяжелого” метана 13СD4 в изобарном процессе при 300 К. Предполагается, что чистый метан доступен в неограниченных количествах.

Решение. Необходимо найти равновесную долю сверхтяжелого метана.
В равновесии доля mС(Dn)H4-n пропорциональна коэффициентам в произведении биномов Ньютона

(0,9888.12С + 0,0112.13С). (0,9998.H + 1,6.10–4.D)4

при соответствующих степенях изотопов, т. е. для 13СD4:

x(13СD4) = 0,0112.1,64.10–16 = 0,0112.6,55.10–16 = 7,34.10–18.

Минимальная работа по выделению 1 моля 13СD4 составит

δWmin = – RT ln(x(13СD4)) = 8,314.300.39,45 = 98,4 кДж моль–1



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