Chem 420/523 Chemical Thermodynamics Homework Assignment # 2 1. Derive explicit expression for the reversible work of isothermal expansion done in each of the following cases: (a) dV is obtained from the equation of state pV = RT + Bp + Cp2 . (b) dV is obtained from the Berthelot equation [Eq. (4.12) on p. 162]. Answer (a) From the equation of state given, RT + B + Cp. p µ ¶ RT dVm = − 2 + C dp, at constant T . p Vm =
Therefore, the work done during an isothermal reversible expansion is w=−
Z
2
1
Z
pdVm = −
Z
Z
2
2
1
µ
p −
¶
RT + C dp p2
2 dp pdp =RT −C 1 p 1 µ ¶ ´ p2 C³ 2 − p2 − p21 . =RT ln p1 2
(b) The Berthelot equation is
"
9pTc pVm = RT 1 + 128pc T
Ã
6T 2 1 − 2c T
!#
.
Expanding and rearranging, we get RT 54RTc3 9RTc − . + p 128pc 128pc T 2 RT dVm =− 2 dP, at constant T . p Vm =
Therefore, the work done during an isothermal reversible expansion is w=−
Z
1
2
pdVm = RT µ
¶
p2 . =RT ln p1
Z
1
2
dp p
2. Rozen [J. Phys. Chem. (USSR) 19, 469 ³ (1945); ´ ³Chem. ´ Abstracts ³ 40, ´ 1712 (1946)] characterizes gases ∂p p ∂V p2 T ∂V by “deviation coefficients” such as p ∂T , R ∂T , and RT ∂p . Calculate these coefficients for V p T (a) ideal gas (b)* van der Waals gas (c)* a gas that obeys the Dieterici equation of state: pVm =
RT −a/(RT Vm ) e . Vm − b
Answer 1
(a) For the ideal gas, µ
¶
·
µ
¶¸
∂p ∂ RT R = = ; ∂T V ∂T Vm V Vm µ ¶ · µ ¶¸ ∂V ∂ RT R = = ; ∂T p ∂T p p p µ ¶ · µ ¶¸ ∂V ∂ RT RT = =− 2 . ∂p T ∂p p p T So, we get µ
¶
T ∂p RT = = 1. p ∂T V pVm µ ¶ p ∂V =1. R ∂T p µ ¶ p2 ∂V =−1. RT ∂p T (b) * For van der Waals’ gas, we have a RT − Vm − b Vm2 µ ¶ ∂p R . = ∂T V Vm − b µ ¶ RT T ∂p = p ∂T V p (Vm − b) a =1+ 2. Vm p=
Since T, p, and Vm are exact differentials, we can use Eq. (1.33) on p. 26 to get an expression for by first calculating
³
∂T ∂V
´
p
:
³
∂V ∂T
´
p
1 a (p + 2 )(Vm − b) R Vm µ ¶ ∂T 2a 1 a =− (Vm − b) + (p + 2 ) 3 ∂V p RVm R Vm 3 pV − aVm + 2ab = m . RVm3 T =
Therefore, µ
∂V ∂T
¶
p
=
"µ
∂T ∂V
p R Also, by the same approach, we calculate above):
³
µ
¶ #−1
=
∂V ∂T
=
p
∂V ∂p
´
T
µ
¶
RVm3 pVm3 − aVm + 2ab
pVm3 . pVm3 − aVm + 2ab
p
by first calculating ∂p ∂V 2
¶
T
=−
³
∂p ∂V
´
T
(see the expression for p
RT 2a 2 + V3. (Vm − b) m
µ
¶
Vm3 (Vm − b)2 RT Vm3 − 2aVm2 + 4aVm b − 2ab2 T µ ¶ p2 ∂V p2 Vm3 (Vm − b)2 = . RT ∂p T RT (RT Vm3 − 2aVm2 + 4aVm b − 2ab2 ) ∂V ∂p
=
(c) * From the Dieterici equation of state, RT e−a/(RT Vm ) Vm (Vm − b) µ ¶ ∂p R a R = × e−a/(RT Vm ) + e−a/(RT Vm ) ∂T V Vm (Vm − b) RT Vm Vm (Vm − b) p a = (1 + ). T RT Vm µ ¶ T ∂p a = (1 + ). p ∂T V RT Vm p=
In this case, it is not possible to write a closed ³ form ´ expression for T . Therefore, we directly differentiate the equation given to get an expression for ∂V : ∂T p
³
´
"
µ
¶
R ∂Vm RT RT a = + + − p 2 (Vm − b) (Vm − b) ∂T# p (Vm − b) RT 2 Vm µ ¶ RT ∂Vm a e−a/(RT Vm ) (Vm − b) RT Vm2 ∂T p µ ¶ µ ¶ µ ¶ ∂Vm ∂Vm ∂Vm pVm pVm ap ap − p = + + . ∂T p T (Vm − b) ∂T p RT 2 RT Vm ∂T p ∂Vm ∂T p
Dividing both sides by p and collecting terms containing the partial derivative, we get µ
¶µ
¶
∂Vm Vm Vm a a 1+ = . − + (Vm − b) RT Vm ∂T p T RT 2 µ ¶ µ ¶µ ¶−1 p ∂Vm pVm Vm ap a = 1+ . + 2 2 − R ∂T p RT R T (Vm − b) RT Vm Proceeding in a similar manner for
³
∂V ∂p
´
T
, we get
RT e−a/(RT Vm ) p (Vm − b) " µ ¶ µ ¶ µ ¶ # ∂Vm ∂Vm ∂Vm RT a RT RT − = − 2 + e−a/(RT Vm ) ∂p T p (Vm − b) p (Vm − b)2 ∂p T p (Vm − b) RT Vm2 ∂p T · µ ¶ µ ¶ ¸ ∂Vm a ∂Vm Vm Vm = − + Vm . − p (Vm − b) ∂p T RT Vm2 ∂p T Vm =
Collecting terms containing the partial derivative, we get ·
¸µ
¶
∂Vm Vm Vm a 1+ =− − . (Vm − b) RT Vm ∂p T p µ ¶ · ¸−1 p2 ∂Vm a pVm Vm − =− 1+ . RT ∂p T RT (Vm − b) RT Vm 3
3. Calculate w, ∆U, q, and ∆H in an isothermal expansion of 1 mole of a gas that obeys the ³ ´ reversible ³ ´ ∂p ∂U = T ∂T − p.] equation of state pV = RT + Bp. [Hint: ∂V T
V
Answer
From the equation of state, we have p(Vm − B) = RT or p = w=−
We are given
³
∂U ∂V
´
T
=T
³
∂p ∂T
´
V
Z
2
1
Z
RT . Therefore, Vm − B
2
RT dVm 1 Vm − b ! Ã Vm,2 − B = −RT ln . Vm,1 − B
pdVm = −
− p. From the equation of state,
µ
∂U ∂V
¶
T
µ
∂p ∂T
¶
=
V
R Vm − B
RT = − p = 0. Vm − B
Therefore, ∆U = 0 for an isothermal expansion process. Since ∆U = 0, using the first law, we get Ã
!
Vm,2 − b . q = −w = RT ln Vm,1 − b The enthalpy change is dH = d(U + pVm ) = dU + d(pVm ) = 0 + d(RT + Bp). Since the process is isothermal, dH = Bdp. Therefore, ∆H = B(p2 − p1 ). 4. A gas obeys the equation of state pV = RT + Bp and has a heat capacity CV,m that is independent of temperature. (a) Derive an expression relating T and V in an adiabatic reversible expansion. (b) Derive an equation for ∆H in an adiabatic reversible expansion. (c) Derive an equation for ∆H in an adiabatic free expansion. Answer (a) For an adiabatic process, dq = 0. Therefore, from the first law, we have dU = −pdVm , or CV,m dT = −pdVm . Substituting for p from the equation of state, this becomes CV,m dT = −
RT dVm . Vm − B
This equation can be rearranged and both sides integrated as follows: dT dVm =R TÃ Vm −!B µ ¶ T2 Vm,2 − B = R ln . CV,m ln T1 Vm,1 − B CV,m
4
Therefore, we get T2 = T1
Ã
Vm,2 − B Vm,1 − B
!R/CV,m
.
(b) From the definitions, dH = d(U + pVm ) = CV,m dT + d(RT + Bp) = CV,m dT + RdT + Bdp. Thus, dH = (CV,m + R)dT + Bdp. [We cannot use the relationship that CV,m + R = Cp,m because this is not an ideal gas!] Since CV,m is independent of tempreature (given) and R is a constant, we get ∆H = (CV,m + R) (T2 − T1 ) + B(p2 − p1 ). (c) In a free expansion, no work is done and so, dw = 0. Also, by definition, dq = 0. Therefore, from first law, we get dU = dq + dw = 0! Since dU = CV,m dT by definition, this implies that dT = 0 for this process. In other words, the temperature remains constant. Therefore, dH = dU + d(pVm ) = dU + RdT + Bdp = Bdp, ∆H = B(p2 − p1 ). h
5. * Given that Cp,m = CV,m + V − or (1.38); p. 29], show that
³
´ i³
∂H ∂p T
∂p ∂T
´
V
·
CV,m = Cp,m 1 − µJT
, and the chain rule of partial derivatives [Eq. (1.37) µ
∂p ∂T
¶ ¸
−V
µ
¶
−V
µ
V
∂p ∂T
¶
.
V
Answer From the given equation, CV,m = Cp,m +
µ
∂H ∂p
¶ µ T
∂p ∂T
V
∂p ∂T
¶
.
V
Since enthalpy s a function of T and p, i.e., H(T, p), we write dH =
µ
∂H ∂p
¶
T
µ
∂H dp + ∂T
¶
dT. T
At constant enthalpy (as in the Joule-Thompson experiment), dH = 0 and we get µ
¶
µ
¶
∂H ∂H dp = − dT ∂p T ∂T T µ ¶ µ ¶ µ ¶ ∂H ∂H ∂T =− ∂p T ∂T T ∂p H = −Cp,m µJT . Substituting in Eq. (1), we get ·
CV,m = Cp,m 1 − µJT
µ
5
∂p ∂T
¶ ¸ V
−V
µ
∂p ∂T
¶
V
.
(1)
6. * From the hint given in Question 3 and the result derived in Question 2(a) of Homework Assignment #1, show that Cp,m − CV,m = α2 V T /κ, where α and κ are defined in Chapter 1 [Eqs. (1.40) and (1.39), respectively]. Answer From Q. 3, we have
µ
∂U ∂V
From Q. 2(a) of assignment 1, we have µ
Substituting for
³
´
∂U ∂V T
∂H ∂T
¶
µ
¶
µ
=T
T
¶
∂U ∂T
∂p ∂T
·
¶
V
µ
(2)
− p. ¶ ¸µ
¶
∂U ∂V = + p+ . ∂V ∂T p p V T · µ ¶ ¸µ ¶ ∂U ∂V . Cp,m = CV,m + p + ∂V T ∂T p
(3)
from Eq. (2), we get ·
Cp,m = CV,m + p + T or Cp,m − CV,m = T Now, by definition,
µ
∂p ∂T
¶
µ
∂p ∂T
¶ µ
µ
V
V
¸µ
∂V ∂T
−p
∂V ∂T
¶
.
p
¶
1 ∂V V ∂T p µ ¶ 1 ∂V κ=− V ∂p T α=
Using the chain rule of partial derivatives (see Eqs. (1.43) and (1.44)), µ
∂p ∂T
¶
= V
Substituting in Eq. (4), and recognizing that
µ
− (∂V /∂T )p α = . (∂V /∂p)T κ
∂V ∂T
¶
= αV , we get
p
Cp,m − CV,m = α2 V T /κ.
6
¶
,
p
(4)