Department
of Chemistry
CH
431 Mid-term I
Given: R = 8.314 J mol-1
K-1 = 0.08206 L atm mol-1 K-1
1 a.m.u. = 1.672 x 10-27 kg
1 atm = 1.0133 x 105
Nm-2 = 760 Torr
P = P0exp
1.
A. The brown color in smog arises from N2O4
that exists in equilibrium with NO2 (2 NO2 ßà N2O4). Assuming that NO2 is produced at
levels of 2 parts per million in
r(gL-1)
= ___________________ (10 points).
At 1 ppm the partial pressure of N2O4
is 2 x 10-6 atm. Therefore the density is:
B.
One researcher decides not to use a van der Waal's gas model to improve on the estimate obtained part
1. A. The research team leader disagrees and asks the researcher to do the
calculation. Is the research team leader
making a sensible decision? (5 points).
The van der Waal's
gas is needed for dense gases. Even at 1
atmosphere of pressure the van derWaal's gas provides
a small correction. At
a pressure of 10-6 atm. The van der Waal's correction will be
very small. The research tem leader is
wasting the researcher's time.
C.
How does the average velocity of N2O4 molecules at 300 K
compare with the velocity of atmospheric nitrogen?
Average
velocity of N2O4 = _____899___________m/s (8
points).
Average
velocity of N2 = ________1634_________ m/s (7 points).
The velocity can be calculated from the kinetic theory of gases.
The ratio of the velocities if
provided by the inverse ratio of the square roots of the molar masses.
Of course you could just solve directly for the velocity of N2O4
as well.
2.
A. Given the partition function for a
hard sphere gas calculate the equation of state.
In
this expression B is the hard sphere volume of one gas molecule. The conventional term b is the volume of one
mole of gas molecules (close packed).
Thus, nb = NB.
You will need to use the expression for the
average pressure (10 points).
Solution:
B. Calculate the average energy of a hard
sphere gas. How does this energy differ
from that of an ideal gas? (10 points).
Solution:
The average energy is the same as for an ideal gas.
3. The core of Jupiter consists of liquid metallic
hydrogen with a radius of approximately 10,000 km. This exotic form of the most common of
elements is possible only at pressures exceeding 4 million bars. Fluid metallic
hydrogen consists of ionized protons and electrons (like the interior of the Sun
but at a far lower temperature). At the temperature and pressure of Jupiter's
interior hydrogen is a dense fluid. The
outer portion of the planet consists of a dense gas that is mostly hydrogen, H2. The radius of Jupiter is usually quoted as
the radius at which the pressure of the gas is 1 atmosphere.
A.
Assuming that the average temperature is
300 K due to the heat of the liquid core and that the average gravitational
acceleration is 0.31 m/s2 and the pressure at the surface of the
liquid core is 4 million bars please calculate the radius of the "gas
planet" Jupiter. Include the 10,000
km of inner core in your answer (15 points).
Solution: The height above the core
is
When added to the radius of the dense core the outer radius is
71,000 km. This is the experimental
value as well.
B.
Assuming that the mass of the core of
metallic liquid hydrogen is 2.1 x 1025 kg calculate the density of
the core (5 points).
r = _____5.0_______________
g/cm3
Solution:
C.
Assuming that the compressed liquid core
of Jupiter can be treated as a hard sphere gas at its maximum density,
calculate the parameter b for metallic hydrogen gas (10 points).
b
= ______________________ L/mol.
Starting with the expression for a hard sphere gas:
At very large pressures
4.
Treating the CO2 gas in the
cylinder of an engine as an ideal gas that is initially at 600 atm and 600 K following combustion, calculate the work done
and heat transferred for an isothermal expansion under the following
conditions. Its volume is initially 0.02
L and the gas expands to 0.2 L under the following conditions:
(1.)Pexternal
= 0
(2.)Pexternal
= 60 atm
(3.)Pexternal
= Pgas
(reversible expansion)
For each of the above conditions calculate DU
and w for the gas (30 points).
q(J) w(J)
(1.)
____0______ _____0______
(2.) __1094 J___ ___-1094 J____
(3.) __2800 J___ ___-2800 J____
1.
An expansion against zero
pressure does no work and w = 0. For an
isothermal process DU = 0. Thus, q = 0 as well.
2.
For an expansion against a
constant pressure
w = -Pext(V2
- V1) = 60 atm (0.2 - 0.02)L = -10.8 L-atm = - 1094 J
Since DU = 0 here too we have q = -w = +
1094 J.
3.
For a reversible expansion
Again q = -w.
5.
A molecule has the three level energy
diagram shown below.
A.
Based on the energy levels shown in the
Figure, write an expression for the molecular partition function assuming that
the particle is fixed in space (i.e. it cannot translate, rotate etc.). (5
points)
B.
Assume that the particle can translate
and rotate, write down an expression for its molecular partition function q. (5
points)
C.
Assuming that a system consists of N
indistinguishable particles in an ideal gas write down an expression for the
system partition function. (5 points)
D.
Write an expression for the average energy
of a system of N particles including translation, rotation and the energy
levels shown in the Figure. Express your answer in terms of e
and b or kT. Assume that e
> kT. (15
points)
Solution:
Only terms in b survive.
E.
Write an expression for the heat capacity
for the system as a function of temperature.
Assume that e
>> kT and use the most convenient approximation
for the electronic levels (10 points).
If e >> kT then only the lowest electronic level is significantly
populated. The energy in this level is e = 0 and this level does note
contribute to the heat capacity. In
other words,
F.
Derive an equation for the pressure of
the system consistent with above assumptions.
(10
points)
Only the translational partition function contributes to the
pressure. A similar problem is solve in
2 A.
6.
The work done by an isothermal expansion
is said to be reversible. To see what
this means we will compare a reversible and an irreversible process. Two systems containing an ideal gas have the
same initial volume of 25 L, pressure of 100 atm, and temperature of 500
K. The systems expand and are then
recompressed along two different paths. The first is one isothermal and the
second one constant pressure. For each
expansion the final volume is 2500 L.
A.
Calculate the work of isothermal
expansion from 25 L to 2500 L and the work of compression from 2500 L back to
25 L.
Isothermal
expansion w = _____-1.166 x 106
J_______ (5 points)
Isothermal
compression w = ___+1.166 x 106
J _______ (5 points)
The isothermal work of expansion is:
The isothermal work of compression is equal in magnitude and
opposite in sign to the work of expansion.
B.
Calculate the work of a constant pressure
expansion from 25 L to 2500 L. Assume that the external pressure is equal to the
final pressure. Then calculate the work
for compression back to 2500 L at constant pressure (assuming the that the pressure is equal to the final pressure).
Constant
pressure expansion w = ____-2.508 x 105
J____________ (5 points)
Constant
pressure compression w = _____+2.508 x 107
J________ (5 points)
The expansion must occur against a final pressure the must be
calculated.
We use P1V1 = P2V2. P2 = (V2/V1)P1 = (25/2500)100 atm = 1atm.
The work of compression must occur against the initial pressure of
100 atm. We must push hard enough to get the piston
back where it started.
C.
How does the work along a reversible and
irreversible path compare? Is work a
state function? (5 points)
Work is a state function. The work of
expansion is maximum along the reversible path. The work of compression is minimum
along a reversible path. The work along
the constant pressure path is 100 times as large for compression as the work
done by the system during the expansion.