2007
Physical Chemistry 2
Homework assignment 1, Solutions
Problem 1:
The coefficient of isothermal compressibility is defined as
1 ∂V
κT = −
V ∂P T
and the coefficient of thermal expansion is defined as
1
α =
V
∂V
∂T
P
Derive expressions for the coefficients of isothermal compressibility and thermal expansion
using the equation of state
(a) for an ideal gas,
(b) given by the expression P (Vm − b) = RT,
(c) of the van der Waals form.
Solution
(a) For an ideal gas, P V = nRT , the volume is given by V = nRT /P , so taking the
partial derivatives gives
κT
1
= −
V
∂V
∂P
1
α =
V
1
= − nRT
V
T
∂V
∂T
=
P
1
1
− 2
=
P
P
1 nR
1
=
V P
T
(b) For a gas with the equation of state P (Vm − b) = RT where Vm = V /n, the molar
volume is given by Vm = RT /P + b. Taking the partial derivative with respect to P gives
κT
1
=−
Vm
∂Vm
∂P
T
1
= −
Vm
RT
Vm − b
1
b
− 2
=
=
−
P
Vm P
P
Vm P
where the correction to the ideal gas result is clearly ∆κT = −b/Vm P .
Similarly, taking the partial derivative with respect to T gives
1
α =
Vm
∂Vm
∂T
P
1
=
Vm
R
P
1
=
Vm
1
Vm − b
T
=
1
b
−
T
T Vm
where the correction to the ideal gas result is clearly ∆α = −b/T Vm . So, both the
compressibility and the thermal expansion are reduced by the deviation of this equation
of state from the ideal gas equation of state.
(c) For a van der Waals gas, the equation of state is (P + a/Vm2 )(Vm − b) = RT . This can
be rewritten to isolate P in the following way
P =
a
RT
− 2
(Vm − b)
Vm
The compressibility coefficient can be written in terms of the partial derivative of P with
respect to Vm
1
1
κT = −
∂P
Vm
∂Vm
T
By evaluating this partial derivative, one gets
κT =
1
Vm
RT
(Vm −b)2
−
2a
3
Vm
Similarly, the thermal expansion coefficient can be written in terms of a partial derivative of T with respect to Vm
1
1
α =
∂T
Vm
∂Vm
P
The temperature, T , van be isolated from the equation of state in the following way
P Vm
ab
Pb
a
−
−
+
R
R
RVm
RVm2
T =
and after differentiating this expression with respect to Vm at constant P , one obtains
α =
1
1
Vm P − 0 − a +
R
RV 2
m
= 2ba
P Vm −
RV 3
m
R
a
Vm
+
2ba
2
Vm
= 1
T+
bP
R
−
2a
Vm R
+
3ab
2R
Vm
which can be rewritten as a cubic equation in the molar volume
P Vm3 − (bP + RT )Vm2 + aVm − ab = 0
By calculating the partial derivative of this equation with respect to P holding T fixed,
one gets
1·
Vm3
+ P·
3Vm2
∂Vm
∂P
−
bVm2
− (bP + RT )2Vm
T
2
∂Vm
∂P
T
+ a
∂Vm
∂P
T
= 0
After dividing by Vm3 this equation becomes
0 = 1 + 3P (−κT ) −
b
2(bP + RT )
a
−
(−κT ) +
(−κT )
Vm
Vm
Vm
After bringing the terms which contain κT to the left hand side, the equation becomes
b
a
bP + RT
= 1 −
+ 2
κT 3P −
Vm
Vm
Vm
Isolating κT gives
κT = 3P −
1 −
b
Vm
bP +RT
Vm
+
a
2
Vm
Similarly, to get the thermal expansion coefficient, calculate the partial derivative of
the equation with respect to T. This gives
α ==
R
3P Vm − 2(bP + RT ) + a/Vm
Problem 2:
The heat of combustion of caffeine was determined by first burning benzoic acid and
then caffeine. In both cases the calorimeter was filled with 466 g of distilled water. When
0.0717 g of benzoic acid, C7 H6 O2 (s), were burned as well as 1.1 cm of the iron wire used to
ignite the sample, the temperature increased from 22.615 ◦ C to 23.487 ◦ C. When 0.0624
g of caffeine, C8 H10 O2 N4 (s), were burned, along with 1.8 cm of the wire, the temperature
increased from 22.714 ◦ C to 23.346 ◦ C. What is the heat capacity of the calorimeter with
and without water, and what is the energy change, ∆U , per mol for the combustion of
caffeine? (The energy change for combustion of benzoic acid is ∆U ◦ = −26.434 kJ/g, the
combustion of the wire gives 9.62 J/cm, and the specific heat of water is 1 cal/g deg).
Solution:
A. Find heat capacity of the calorimeter: Burn 0.0171 g of benzoic acid and 1.1 cm of wire.
This gives temperature rise of ∆T =23.487-22.615 deg = 0.872 deg. The heat released is
qV = 0.0717g · 26434J/g + 1.1cm · 9.62J/g = 1895J + 11J = 1906J
The total heat capacity is
CV = CVbomb + CVprod + CVwater = 1906J/0.872deg = 2186J/deg
Subtract heat capacity of the water, CVwater = 466 g · 1cal/g·deg ·4.184 J/cal = 1950 J/deg,
which means that CVbomb + CVprod = (2186 − 1950) J/deg = 236 J/deg
B. Heat of combustion for caffeine: Burn 0.0624 g of caffeine as well as 1.8 cm of wire.
This gives temperature rise of ∆T =22.714-23.346 deg = 0.632 deg. Heat released due to
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combustion of coffeine is ∆U = ∆Utot - ∆Uwire = 0.632deg ·2186 J/deg - 1.8cm·9.62J/cm
= (1382 - 17)J = 1365 J from this sample.
Per gram of caffeine, the heat released is ∆U = 1365 J/0.0624 g = 21.870 kJ/g.
Per mol of caffeine, the heat released is 21.870 kJ/g · 194 g/mol = 4243 kJ/mol
where 194 is the molecular weight (8 · 12 + 10 + 2 · 16 + 4 · 14 = 194).
Problem 3:
Methane gas is heated from 25 ◦ until the volume has doubled. The pressure is constant
at 1 bar. The variation in molar heat capacity with temperature has been measured and
the results of those experiments can be summarized by
C̄P = a + bT
where a = 22.34 JK −1 mol−1 and b = 48.1 × 10−3 JK −2 mol−1 . Calculate ∆H̄ and ∆Ū
per mol, (you can estimate the final volume temperature assuming ideal gas behaviour).
Solution:
The final volume is twice the initial volume, so V2 = 2V1 . The pressure is constant,
so P1 = P2 . The initial temperature is T1 = 298K. The final temperature can be found
assuming ideal gas behaviour, T2 /T1 = P V2 /P V1 = 2. So, T2 = 2T1 = 596 K and
∆T = 298 K.
∆H =
∂H
∂T
P
dT =
Z
T2
Cp (T )dT =
T1
Z
T2
T1
(a + bT )dT =
b
aT + T 2
2
T2
T1
= 6.657 kJ/mol + 6.407 kJ/mol = 13.06 kJ/mol
Find change in internal energy,
∆U = ∆H − P ∆V = ∆H − R∆T = (13.06 − 2.48)kJ/mol = 10.59 kJ/mol
Problem 4:
(a) One mole of a perfect, monatomic gas expands reversibly and isothermally at 25◦ C
from an initial pressure of 5 atm to 1 atm. Determine the value of q, w, ∆U and ∆H for
this process.
(b) One mole of a perfect, monatomic gas expands reversibly and adiabatically from 5 atm
to 1 atm. The initial temperature is 25◦ C. Determine the final temperature of the gas, as
well as q, w and ∆U .
Solution:
(a) Since the temperature does not change (isothermal) and the gas is ideal, the internal
energy does not change, i.e. ∆U = 0. By definition of the enthalpy H = U + P V
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so ∆H = ∆U + ∆(P V ) = ∆U + ∆(nRT ) = ∆U since neither n nor T changes. So,
RV
∆H = 0. The work done in a reversible isothermal process is w = − V12 Pext dV =
−nRT ln V2 /V1 = nRT ln P2 /P1 = 1 mol · 8.314 J/Kmol · 298 K · ln 1/5 = −3988 J. By
the first law, q + w = ∆U = 0 so q = 3988 J.
(b) For an adiabatic process, q = 0. As the gas expands against the external pressure it
does work and looses energy and, since no heat flows in, the temperature drops. The final
temperature can be found from the equation
T2 = T1
P2
P1
(γ−1)/γ
where γ = CP /CV . For an ideal gas CP = CV + nR and for one mol of monatomic
ideal gas CV = 3R/2. Therefore, γ = 5/3 and (γ − 1)/γ = 2/5. The final temperature
is T2 = 298 K · (1/5)2/5 = 156 K. The change in internal energy gan then be obtained,
∆U = (3/2)R∆T = 1.5 · 8.314 J/Kmol · (298 − 156) K = 1770 J. By the first law,
w = ∆U − q = 1770 − 0 = 1770 J. Finally, the enthalpy change is ∆H = ∆U + nR∆T =
1770 J + 8.314 J/Kmol · (298 − 156) K = 2950 J.
Problem 5:
The constant volume heat capacity of a gas consisting of P4 (g) molecules has been
measured to be 67.15 J/Kmol. The P4 molecules are non-linear.
(a) Assuming ideal gas behaviour, what is the expected heat capacity ratio, γ = CP /CV ,
based on the experimental data?
(b) What value of the heat capacity ratio, γ, would one expect from the high temperature approximations to translational, rotational and vibrational partition functions (the
”equipartition theorem” results)?
(c) Often, the vibrational degrees of freedom are not active at room temperature and are
therefore skipped when the heat capacity is estimated. What is the heat capacity ratio,
γ, one would expect from the high temperature approximations if vibrational degrees of
freedom are skipped? Which is more accurate in this case: (i) including vibration or (ii)
skipping it? Explain briefly the results you obtain for P4 (g).
Solution:
(a) For an ideal gas CP = CV + R so γ = CP /CV = 1 + R/CV = 1 + 8.314/67.15 = 1.124
(b) The molecule is non-linear, so the total number of vibrational degrees of freedom
is 3N − 6 = 6 and the high temperature approximation (equipartition theorem) predicts
CV = CV,tr + CV,rot + CV,vib = (3/2)R + (3/2)R + 6R = 9R and a heat capacity ratio of
γequip = 1 + 1/9 = 1.111
(c) Skipping the vibrational degrees of freedom in the high temperature approximation
gives CV = CV,tr +CV,rot = (3/2)R+(3/2)R = 3R and a heat capacity ratio of γequip−vib =
1 + 1/3 = 1.333. In this case it is more accurate to include the vibrational contribution
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to the heat capacity even though the high temperature approximation is used. This is
because the vibrational frequencies of P4 are very low, the molecule is ’floppy’. At room
temperature the vibrational degrees of freedom are to large extent thermally excited and
contribute to the heat capacity.
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