CHM 3410 - Physical Chemistry 1
First Hour Exam
September 21, 2012
There are five problems on the exam. Do all of the problems. Show your work.
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R = 0.08206 L.atm/mole.K
NA = 6.022 x 1023
.
.
R = 0.08314 L bar/mole K
1 L.atm = 101.3 J
.
R = 8.314 J/mole K
1 atm = 1.013 bar = 1.013 x 105 N/m2
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1. (16 points) Consider a mixture of the gases carbon tetrafluoride (CF4, M = 88.01 g/mol) and carbon
tetrachloride (CCl4, M = 153.82 g/mol), with XCF4 = 0.300. The density of the gas mixture, measured at T
= 300.0 K, is  = 1.46 g/L. Assuming the gas mixture behaves ideally what is p, the total pressure of the
gas mixture?
2. (20 points) A substance undergoes the following three step process illustrated in the diagram below.
Step 1 (A  B) A constant pressure reversible decrease in the volume of the substance.
Step 2 (B  C) A constant volume reversible increase in the pressure of the substance
Step 3 (C  A) An adiabatic reversible expansion of the substance
A table of values for q, w, U, and H for each step in the cycle is given below. Some entries in
the table are missing. Find the values for the missing entries, or briefly explain why the values cannot be
found from the information provided. Clearly indicate the values for each missing quantity (or your reason
why they cannot be found) in your answer.
q
w
U
H
Step 1 (A  B)
- 250. J
+ 500.0 J
UAB
HAB
Step 2 (B  C)
+ 750. J
wBC
UBC
HBC
Step 3 (C  A)
qCA
wCA
UCA
HCA
3. (20 points) The Berthelot equation of state is
p=
RT
- a
(Vm – b) TVm2
(3.1)
where Vm = V/n, and a and b are constants.
a) The internal pressure of a substance, T, is given by the expression
T = (U/V)T = T (p/T)V – p
(3.2)
Find an expression for T for a gas obeying the Berthelot equation of state. Simplify your result as much
as possible.
b) Find a general expression for w (work) when 1.000 mol of a gas obeying the Berthelot equation
of state is expanded isothermally and reversibly from an initial volume V i to a final volume Vf.
4. (24 points) Consider the adiabatic expansion of 1.000 mol of an ideal gas from an initial pressure p i =
2.500 atm to a final pressure p f = 1.000 atm. The initial and final temperatures for the gas are T i = 340.0 K
and Tf = 280.0 K. Over this temperature range the constant volume molar heat capacity is C V,m = 20.78
J/mol.K, and can be assumed constant.
Find q, w, U, H, Ssyst, Ssurr, and Suniv for the process.
5. (20 points) The transition metal zirconium (Zr) forms several different gas phase compounds with
bromine. A partial table of thermodynamic data for some of these compounds, at T = 25.0 C, is given
below, and can be used to answer the questions in this problem.
Substance
M (g/mol)
Hf (kJ/mol)
S (J/mol.K)
Cp,m (J/mol.K)
Zr(s)
ZrBr2(g)
ZrBr4(g)
91.2
251.0
410.8
0.00
- 174.47
- 644.75
38.9
316.8
414.5
25.20
60.68
102.67
Consider the reaction
Zr(s) + ZrBr4(g)  2 ZrBr2(g)
(5.1)
a) What are Grxn, Hrxn, and Srxn for reaction 5.1 at T = 25.0 C?
b) Is reaction 5.1 spontaneous for standard conditions at T = 25.0 C (yes/no and a justification for
your answer)?
c) What is Srxn for reaction 5.1 at T = 100.0 C?
Solutions.
1) Assume you have 1.000 mol of gas.
Then
nCF4 = 0.300 mol
nCCl4 = 0.700 mol
The mass of gas is
m = mCF4 + mCCl4 = nCF4 MCF4 + nCCl4 MCCl4
= (0.300 mol)(88.01 g/mol) + (0.700 mol)(153.82 g/mol) = 134.08 g
The volume of gas is then
V = m = 134.08 g = 91.83 L
 1.46 g/L
For an ideal gas, or a mixture of ideal gases
p = nRT = (1.000 mol)(0.082057 L.atm/mol.K)(300.0 K) = 0.268 atm ( = 204. torr)
V
91.83 L
2)
Step 2 is constant volume, and so wBC = 0.
Step 3 is adiabatic, and so qCA = 0.
From the first law, U = q + w
So
UAB = qAB + wAB = (- 250. J) + (500. J) = + 250. J
UBC = qBC + wBC = (+ 750. J) + (0. J) = + 750. J
Since U is a state function
UAB + UBC + UCA = 0
UCA = - (UAB + UBC) = - (250. J + 750. J) = - 1000. J
UCA = qCA + wCA
and so
wCA = UCA – qCA = (- 1000. J) - (0. J) = - 1000. J
Step 1 is constant volume, and so HAB = qAB = - 250. J
Since H is a state function
HAB + HBC + HCA = 0
However, we have no way to determine either HBC + HCA, and so values for these two quantities cannot
be found.
3)
a) To find the internal pressure we need to evaluate the partial derivative (p/T)V
(p/T)V = /T)V
RT
- a
=
R
+ a
(Vm – b) TVm2 (Vm – b) T2Vm2
T = T (p/T)V – p =
RT
+
(Vm – b)
RT
+
a
(Vm – b)
TVm2
- p
a
- RT
+
a
TVm2
(Vm – b)
TVm2
= 2a
TVm2
b) The process is isothermal and reversible, and so
w = - if pex dV = - if p dV
= - if { [nRT/(V – nb)] - (an/TV2) } dV
Note we have used Vm = V/n to rewrite the expression for p in terms of n and V instead of in terms of V m.
So
w = - { RT ln[(Vf – nb)/(Vi – nb)] + (an/T) [ (1/Vf) – (1/Vi) ] }
4)
The process is adiabatic, and so q = 0 and Ssurr = 0.
Therefore
w = U
Suniv = Ssyst
Since the gas is ideal and the heat capacity constant (and noting that Cp,m = CV,m + R = 29.09 J/mol.K)
U = if nCV,m dT = nCV,m if dT = nCV,m (Tf – Ti)
= (1.000 mol)(20.78 J/mol.K)(280.0 K – 340.0 K) = - 1247. J
H = if nCp,m dT = nCp,m if dT = nCp,m (Tf – Ti)
= (1.000 mol)(29.09 J/mol.K)(280.0 K – 340.0 K) = - 1745. J
Finally, to find Ssyst we need a reversible pathway connecting the initial and final state. One such pathway
is
Step 1 – Isothermal reversible expansion of the gas, from pi = 2.500 atm to pf = 1.000 atm, at T =
340.0 K.
Step 2 – Constant pressure cooling of the gas, from T i = 340.0 K to Tf = 280.0 K, at p = 1.000 atm.
We have discussed both of these types of processes in class and so will simply use the results previously
obtained.
S1 = nR ln(pi/pf) = (1.000 mol)(8.3145 J/mol.K) ln(2.500/1.000) = 7.618 J/K
S2 = nCp,m ln(Tf/Ti) = (1.000 mol)(29.09 J/mol.K) ln(280.0/340.0) = - 5.648 J/K
Ssyst = S1 + S2 = (7.618 J/K) + ( - 5.648 J/K) = + 1.97 J/K = Suniv
5)
a) Hrxn = [ 2 Hf(ZrBr2(g)) ] – [ Hf(Zr(s)) + Hf(ZrBr4(g)) ]
= [ 2( - 174.47) ] – [ 0.0 + (- 644.75) ] = + 295.81 kJ/mol
Srxn = [ 2 S(ZrBr2(g)) ] – [ S(Zr(s)) + S(ZrBr4(g)) ]
= [ 2(316.8) ] – [ 38.9 + (414.5) ] = 180.2 J/mol.K
Grxn = Hrxn - TSrxn = 295.81 kJ/mol – (298.15 K)(180.2 x 10-3 kJ/mol.K)
= 242.08 kJ/mol
b) Since Grxn > 0, the reaction is not spontaneous for standard conditions at T = 25.0 C.
c) Srxn(T2) = Srxn(T1) + T1T2 (Cp/T) dT
If we assume the heat capacities are independent of temperature, then
Srxn(T2) = Srxn(T1) + (Cp) ln(T2/T1)
Cp = [ 2 Cp,m(ZrBr2(g)) ] – [ Cp,m(Zr(s)) + Cp,m(ZrBr4(g)) ]
= [ 2(60.68) ] – [ 25.20 + (102.67) ] = - 6.51 J/mol.K
Srxn(100.0 C) = 180.2 J/mol.K + (- 6.51 J/mol.K) ln(373.15/298.15) = 178.7 J/mol.K