PY2104 - Introduction to thermodynamics and Statistical physics ...

PY2104 - Introduction to thermodynamics and Statistical
physics
Review Problems - solutions. Part III: Thermodynamic
functions and equilibrium
1) We know that the temperature T is given by
( )
1 ∂S
T = = n ∂U 2 ×5R 1 U
from which we get that
V
U = 5 2 nRT .
The specific heat at constant volume is thus
( ) dU
c v = = 5 dT 2 nR
The specific heat at constant pressure is
c p = c v +nR = 7 2 nR .
2) (i) We have done this in class. In an adiabatic process, TdS = 0, and hence
c v dT = −pdV. Using the perfect gas equation of state, pV = Nk B T, we get
from which γpdV +Vdp = 0, or
pdV +Vdp = Nk B dT = (c p −c v )dT ,
V
pV γ = const .
(ii) The reason that c p > c v is that the gas expanding at constant pressure
has to do work, and therefore more heat is absorbed for this purpose.
3) (i)
(ii)
dQ = dU +dW = −d(MB)+MdB = −BdM
dS = dQ T = −B T dM = −M C dM .
(iii)
S = S 0 − M2
2C
4) We can use the fundamental relation,
[ ( )
dU +pdV 1 ∂U
dS = = + p T T ∂V
T
T
]
dV + 1 T
( ) ∂U
dT .
∂T
V
Substituting the expressions for U and p, we get
dS = BTn−1 +AT 2 [ f ′ (T)
dV + +nBT n−2 ln V ]dT .
V T V 0

Using ∂ 2 S/∂T∂V = ∂ 2 S/∂V∂T, we get
(
∂ BT n−1 +AT 2 )
= ∂ [ f ′ (T)
+nBT n−2 ln V ]
∂T V ∂V T V 0
,
from which we get
2AT −BT n−2 = 0 .
The solution must therefore be n = 3 and B = 2A.
5) (i) We can write the equation of state as
p = RT
V −b − a
V 2 .
In an isothermal process, the change in Helmholtz free energy is
∆F = − ∫ V 2
V 1
pdV
(
= int
)
V2
( V
dV 2 )
V
= −RT ln 2−b 1
V 1−b
+a
V 1
− 1 V 2
.
RT
V 1 V−b − a
(ii) The change of internal energy can be calculated using
( ) ( ) ∂U ∂U
dU = dT + dV .
∂T ∂V
For an isothermal process, we have
( ) ∂U
dU =
∂V
We can further use
V
( ) ∂U
= T
∂V
T
Using the equation of state, we get
Integrating, we get
∆U =
∫ V2
V 1
T
dV .
T
( ) ∂p
−p
∂T
V
dU = a
V 2dV .
(
a 1
V 2dV = a − 1 )
V 1 V 2
.
6) The entropy is given by
( ) ∂G
S = −
∂T
P
= 5 [ ]
2 R−Rln ap
(RT) 5/2
The specific heat at constant pressure is
( ) ∂S
c p = T = 5 ∂T 2 R .
p
.

7) Let the molar numbers of the gas in the two sides (A) and (B) be n 1 and n 2 ,
respectively. From the equations of state: pV = n 1 RT and 6pV = n 2 RT,
we obtain n 2 = 6n 1 .
(i) The system is isolated and containing an ideal gas, and therefore the
final temperature is T f = T, since in both sides the initial temperature
is T.
(ii) The final pressure is
p f = (n 1 +n 2 )RT
3V
= p 3
(
1+ n )
2
= 7 n 1 3 p .
(iii) The change in entropy can be calculated by designing a quasi-static
isothermal process. Then,
(∫ )
∆S = 1 T p1 dV 1 +p 2 dV 2
= n 1 R ∫ V 1,f dV
V V +n 2R ∫ V 2,f dV
2V V
= n 1 Rln V 1,f
V +n 2Rln V 2,f
V
Since V 1,f +V 2,f = 3V, and V 2,f
V 1,f
= n2
n 1
= 6, we find that V 2,f = 6V 1,f =
18
7 V. Thus, ∆S = n 1 Rln 3 7 +n 2Rln 9 7 ≈ 2 pV
3 T .