Chemistry 5350
Advanced Physical Chemistry
Fall Semester 2013
Problems on Fundamental Concepts of Thermodynamics
Chapter 1
September 7, 2013
1. At sufficiently high temperatures, the van der Waals equation has the form
P =
RT
n2 a
RT
a
nRT
− 2 =
− 2 ≈
V − nb
V
Vm − b Vm
Vm − b
Note the attractive part of the potential has no influence in this expression.
Justify this behavior using the potential energy diagram of Figure 1.10 shown here.
At high temperatures, the energy of the molecule is large and far above the minimum in energy of the
well. The well depth is a fraction of the total energy, and the molecule is unaffected by the attractive
part of the potential.
2. Explain why attractive interactions between molecules in a gas make the pressure less than predicted
by the ideal gas equation.
The ideal gas equation assumes that the molecules in a gas are point particles that don’t interact. But
real molecules do interact and have attractive interactions at moderate pressures and low energies.
This attraction between molecules lowers the pressure with respect to molecules which are assumed
to not interact.
3. A gas sample is known to be a mixture of ethane and butane. A bulb having 230.0 cm3 capacity is
filled with the gas to a pressure of 97.5 × 103 Pa at 23.1 ◦ C. If the mass in the bulb is 0.3554 g, what
is the mole percent of butane in the mixture?
n1 = moles of ethane; n2 = moles of butane.
ntotal = n1 + n2 =
PV
97.5 × 103 Pa × 0.230 × 10−3 m3
=
= 9.106 × 10−3 mol
−1
−1
3
RT
8.314 Pa m mol K × 296.2 K
n1 M1 + n2 M2 = 0.3554 g
n 2 M2
0.3554 g
n 1 M1
= 39.03 g mol−1
+
=
n1 + n2 n1 + n2
9.106 × 10−3 mol
x1 M1 + x2 M2 = (1 − x2 )M1 + x2 M2 = 39.03 g mol−1
39.03 g mol−1 − 30.069 g mol−1
39.03 g mol−1 − M1
=
x2 =
M2 − M1
58.123 g mol−1 − 30.069 g mol−1
4. Consider the reaction of the amino acid glycine NH2 CH2 COOH to produce water carbon dioxide,
and urea, NH2 CONH2 :
2NH2 CH2 COOH(s) + 3O2 (g) −−→ NH2 CONH2 (s) + 3CO2 (g) + 3H2 O(l)
Calculate the Volume of carbon dioxide evolved at P = 1.00 atm, and T = 305 K in the oxidation
of 0.022 g of glycine.
1 mol NH2 CH2 COOH ≎
nglycine =
3 mol
CO2
2 mol
0.022 g
0.022 g
≎3×
= nCO2
75.07 g/mol
75.07 g/mol
nCO2 × 8.206 × 10−2 L atm mol−1 K−1 × 305 K
nCO2 RT
=
V =
P
1.00 atm
5. A mixture of H2 and NH3 has a volume of 139.0 cm3 at 0.00◦ C and 1 atm. The mixture is cooled to
the temperature of liquid nitrogen at which ammonia freezes out, and the remaining gas is removed
from the vessel. Upon warming the vessel to 0.00◦ C , the volume is 77.4 cm3 . Calculate the mole
fraction of NH3 in the original mixture.
Initially, V = 0.1390 L, T = 273.1 K, and P = 1.00 atm
ntotal = nNH3 + nH2 =
PV
RT
After NH3 freezes, H2 is removed, and then warmed to 0.00◦ C , V = 0.0774 L, T = 273.1 K, and
P = 1.00 atm
PV
nNH3 =
RT
xNH3 =
nNH3
ntotal
6. A glass bulb of 0.198 L contains 0.457 g of gas at 759.0 Torr and 134.0◦ C . What is the molar mass
of the gas?
RT
M =g×
PV
M = 0.457 g ×
62.36 L Torr mol−1 K−1 × 407.1 K
759.0 Torr × 0.198 L
7. Rewrite the van der Waals equation using the molar volume rather than V and n.
P =
nRT
n2 a
RT
a
− 2 =
− 2
V − nb
V
Vm − b Vm