CHEM 331
Physical Chemistry
Fall 2013
Name: Answer Key
Midterm Examination 1
1.
A goat weighing 20 kg produces about 375 cm3 gas per hour in his intestines. Assume that
the gas is Methane; CH4 with M = 16 g/mol. How long does it take the goat to produce an
amount of intestinal gas equal to his own weight? Make appropriate approximations.
Explicitly state your approximations.
20000g / (16g/mol) = 1250 mol Methane
V = nRT/P = (1250 mol) (0.08206 Latm/Kmol) (300K) / ( 1atm)
= 30772 L
# hours = 30772 L / (0.375 L/hr) = 82060 hours
Assume: Gas is Ideal and at 1 atm and 300K
2.
A mixture of 1 mole H2 and 2 mole O2 is at 25oC in a 20 L flask. Calculate PH2 and xH2.
Provide a reasonable definition of the partial volume Vi and use it to calculate VH2.
P = nRT/V = (3 mol) (0.08206 Latm/Kmol) (298 K) / (20 L)
= 3.67 mol
xH2 = 1 mol / (1 mol + 2 mol) = 0.333
Dalton's Law:
PH2 = xH2 P = (0.333) (3.67 atm) = 1.22 atm
VH2 = xH2V = (0.333) (20 L) = 6.66 L
3.
The Compressibility Factor Z for N2 gas is 0.85 at a concentration of 10.0 mole/L at 200K.
a) What Pressure would be exerted by this gas, under these conditions, if it is behaving
Ideally?
P = RT /
= RTc = (0.08206 Latm/Kmol) (200 K) (10.0 mol/L)
= 164 atm
b) What Pressure is exerted by this gas, under these conditions?
P = ZRT /
= (0.85) (0.08206 Latm/Kmol) (200 K) (10.0 mol/L)
= 140 atm
c) What is the percentage error introduced into the Pressure calculation by assuming the
gas is Ideal?
% error = |140 - 164| / 140 x 100 = 18%
4.
Suppose that you have data giving the Compressibility Factor (Z) for a gas as a function of
Temperature and Concentration (c); c = 1/ .
a) Derive an expression for calculating
P = ZRT /
from this data.
= ZRTc
b) The Virial expansion for a gas that can be modeled as a van der Waals gas is:
Z =
Develop an expression to determine
for a van der Waals gas that is good through
first order in c.
c) Determine the value of this+ derivative for a van der Waals gas whose constants are:
a = 5.499 L2atm/mol2 and b = 0.0638 L/mol. The temperature of the gas is 273.15K
and its concentration is 0.04464 mol.
= (0.08206 Latm/K) (273.15 K) + 2 (0.08206 Latm/Kmol)
+ 2 (0.04464 mol/L) (0.08206 Latm/K mol\)(0.0638 L/mol
x (0.0638 L/mo - 5.499 L2atm/(0.08206 Latm/Kmole)(273.15K)
= 22.414 L/mol - 0.363 L/mol
= 22.051 Lamt/ mol
5.
The Compressibility Factor for a number of gases as a function of the reduced state
variables is given as:
The critical isotherm for CO2 is as plotted below:
Based on the form of this isotherm, explain why the curve for TR = 1.00 in the above
Z vs. PR diagram is nearly vertical below PR ~ 1.00.
Along the critical isotherm; TR = 1.00 and PR ~ 1.00 at
. You should notice
that near , the isotherm is very flat. This means P/V ~ 0. Along the isotherms of
the Z vs P plot will have a slope of ~ (PV)/ P ~ V/P ~ 1 / 0 = ∞.
6.
For 1 kg of Water between 0oC and 40oC, the volume can be represented as:
V = 999.87 - 0.06426 t + (8.5045 x 10-3) t2 - (6.79 x 10-5) t3
Calculate the Isothermal Compressibility  for Water at 3oC. Recall:
 =
Do you consider this result unusual? Comment Briefly.
V = 999.87 - (0.06426)(3) + (8.5045 x 10-3) (3)2 - (6.79 x 10-5) (3)3
= 999.75
= -0.06426 + 2 (8.5045 x 10-3) t -
3 (6.79 x 10-5) t2
= -0.06426 + 2 (8.5045 x 10-3) (3) -
3 (6.79 x 10-5) (3)2
= - 0.01507
So,
 = - 0.01507 / 999.75 = - 1.507 x 10-5 K-1
This value is unusual in that it is negative. This means the Water is contracting
when heated.
7.
A resistor is immersed in a bath of crushed Ice in a Dewar flask. A current of 2 Amperes
passed through the resistor for 300 sec (5 min) results in the melting of some of the Ice.
The resistance is 18 . (Recall, the electrical work done on a resistor in passing a current
through it is given by W = I2Rt.)
a) Find Q, W and U for the System. Take the System to be the Ice-Water.
QWater = - QResistor = (2 Amperes)2 (18 ) (300 sec) = 21600 J
W ~ 0J
U = Q + W = 21600 J + 0 J = 21600 J
b) Suppose all the Ice melts before the heating is complete. Will that affect your
calculations above? Explain.
If the Ice melts before the heating is complete, then the Water will warm.
This will leave the resistor in a different state than it started in; meaning U
is no longer zero for the resistor. Thus, some of the work on the resistor will
be used in changing U. Hence we do not know how much heat is delivered to
the Ice-Water. Hence we cannot calculate Q or U for the Ice-Water.
8.
One mole of an Ideal Gas at 2 atm & 0oC undergoes reversible Isobaric heating until it is at
2 atm & 50oC. This then undergoes a reversible Isothermal compression to 2.1 atm.
Calculate W for each leg of the path in going from state A to state B. Clearly identify,
for each leg of the path, whether work is being done on the system by the surroundings,
or vice versa.
VB = nRT/P = (1 mol) (0.08206 Latm/Kmol) (323 K) / (2.1 atm) = 12.6 L
Along Leg #1 of the Path:
W = -
= -
= -P
= - P V
= - (2 atm) (13.3 L - 11.2 L) = - 4.2 Latm
Work is being done by the system on the surroundings.
Along Leg #2 of the Path:
W = -
= = - nRT
=
-
= - nRT
= - (1 mol) (0.08206 Latm/Kmol) (323 K)
= + 1.4 Latm
Work is being done by the surroundings on the system.