На главную Институт катализа им. Г.К. Борескова СО РАН
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Версия для печати | Главная > Образование > Кафедра физической химии > Химическая термодинамика > ... > Задачи по химической термодинамике > Решения некоторых задач

1. Основные понятия химической термодинамики

Переход к разделу

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 )(Vb)=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=TdSPdV 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 ( Vb ) a V 2 ) T ) V = R ( Vb ) = 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 TP 2 A TV MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaaBaaaleaacaWGWbaabeaakiabgkHiTiaadogadaWgaaWcbaGaamOvaaqabaGccqGH9aqpcqGHsislcaWGubWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGhbaabaGaeyOaIyRaamivaiabgkGi2kaadcfaaaWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGbbaabaGaeyOaIyRaamivaiabgkGi2kaadAfaaaaaaa@4D10@

Решение.Cначала упростим выражение:
2 G TP = ( V T ) P , MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITdaahaaWcbeqaaiaaikdaaaGccaWGhbaabaGaeyOaIyRaamivaiabgkGi2kaadcfaaaGaeyypa0ZaaeWaaeaadaWcaaqaaiabgkGi2kaadAfaaeaacqGHciITcaWGubaaaaGaayjkaiaawMcaamaaBaaaleaacaWGqbaabeaakiaacYcaaaa@4676@ а 2 A TV = ( 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 8,314298 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 ЖТ 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 =8,314268,2ln 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

2 (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.


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