Основные понятия химической термодинамики (решение задач)
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.