HW #2:
Kinetic Theory of Gases
Classical Physics
Assigned: 2/05/09
Due: 2/12/09
Introductory Modern Physics
(95.210)
Spring 2009
Instructor: Prof. Wasserman
Olney 126
daniel_wasserman@uml.edu
(978) 934-4530
Relativity
1) Prof. Wasserman gives his first mid-term exam. After grading all but
the last exam, the distribution looks like:
2 students = 13
4 students = 29
1 students = 54
3 students = 66
1 students = 72
3 students = 78
2 students = 82
a) Professor Wasserman was hoping for the average grade squared to be
at least 3080. What does the grade on the final exam need to be to bump
the average grade squared over 3080? Use the expression for finding the
average value that we used in class.
b) Assuming the average score squared is 3080, what is the standard
deviation of this distribution?
a) The average score for the class is given by the formula:
s
1
N
s n
i
i
i
Where N is the number of students in the class: 17
si is a given score
ni is the number of students with score si.
The average grade squared is simply s 2
We can do the sun above to get:
1
(864  finalscore )
17
s  3080
s
finalscore  17 * 3080  864
finalscore  79.46
b) If the average score squared is 3080, we can calculate the standard
deviation using the formula:
1
si2 ni  3699.05

N i
2
So   3699.05  3080  619.06
  24.88
 2  (s 2 )  s 2 and s 2 
2) A gas is held at 435K. If the gas contains 2.1 Mol of molecules, what
is the average kinetic energy of each molecule in the gas? If the gas is
compressed to half of the original volume, but the temperature is held
constant, what is the change in pressure? What is the average kinetic
energy of each molecule now?
At 435K, the average energy of the molecules in the gas is given by
E
3
kT so E  1.5 *1.38  10 23 * 435  9.0045  10 21 J  0.056eV
2
If the gas is compressed to half the original volume, we can use the ideal
gas law PV  nRT to see that the pressure must increase by a factor of 2.
However, because the temperature has not changed, the average energy
of each molecule will also not change. The number of Mol contained in
the volume is irrelevant.
3) Two different gases are held in a container of fixed volume, and the
temperature and pressure of the gas are monitored. The temperature
is dropped by 30K, and a pressure drop of 5% is seen. The
Temperature is then decreased an infinitesimal amount, and gas 1
condenses. The pressure then drops to 150,000 N/m2 (Pa), a drop of
33%. What is the temperature for the gas->liquid phase transition of
gas 1, for the given pressure and volume? What is the original
pressure of the system, before the temperature change? Can you
determine the molar ratio of the two gases (when both are in gas
phase)?
From the ideal gas law, we know that the change in pressure and
temperature must be proportional. This means that a 30K change in
temperature is also a 5% reduction in temperature. The initial
temperature is then 600K and the reduced temperature must be 570K.
Thus, 570K would be the temperature for the gas-liquid phase transition
of gas 1.
When the first gas condenses, only one gas is left in the volume, and the
pressure drops 33% (this is in addition to the 5% reduction from 600K to
570K). If the final pressure is 150,000 Pa, the initial pressure must be
Pi = 150,000*1.33*1.05 = 236,842 Pa
Finally, the molar gas ratio can be easily calculated. Because P is
proportional to n, when the first gas condenses, and P decreases by 33%,
so must n. So for every molecule of gas 1, there will be two of gas 2.
4)
Calculate vrms for N2 at T=300K.
3
1 2
E  kT  mvrms
2
2
3kT
3RT
v rms 

m
M
we know k and T, but what is M? M is the Molar Mass, or the mass (in
g) of 1Mol of N2. We know that 1mol of N2 weighs 28g (two times the
molar mass of Nitrogen).
From this we conclude that
v rms  517 m / s
5) Calculate Cv for the following situations:
(a) A monatomic gas free to move in three dimensions
Here, each atom can have only translational motion (kinetic energy). IT
can move in three dimensions, so each dimension will give an average
energy of 1/2kT. In total,
E
3
kT and C v  dU
dT V
2
and U  N A E
and
k
R
, so CV=(3/2)R
NA
(b) A monatomic gas free to move in only two dimensions
If the gas can only move in 2D, then it has only two degrees of freedom
and an average kinetic energy of E  kT making CV=R.
(c) A diatomic gas confined to one-dimensional motion
A diatomic gas confined to 1D motion has only one degree of
translational freedom. In addition, because it can only move in 1D, no
rotation is allowed. However, the diatomic molecule can vibrate, so we
get 2 degrees of freedom from the molecular vibration (one for the
kinetic energy component, one for the potential energy, as discussed in
class). Thus, the total average energy comes from 3 degrees of freedom
(1 translational, 2 vibrational). The heat capacity will be the same as for
part (a), CV=(3/2)R.
(d) A diatomic gas free to move in three dimensions
A diatomic gas free to move in 3D has 3 translational degrees of
freedom, 2 rotational, and 2 vibrational. The total is then 7, so
E
7
kT and CV=(7/2)R
2
6) OEQ: Describe an experimental set-up to determine the atomic weight
of table salt (NaCl) and/or Na and Cl atoms individually.
In order to determine the individual atomic weight of NaCl, one must be
able to know the total weight of NaCl. One way to do this would be to
take a NaCl crystal of known mass and volume. The spacing between
molecules could be found by x-Ray diffraction. This would give an
approximate atomic density for the crystal, as well as an average mass for
each molecule.
The crystal could then be placed in water between two electrodes, upon
application of a voltage, the NaCl will separate and Na will deposit on
the negative electrode. By measuring the weight of the electrode before
and after the voltage is applied, one can determine the mass of the Na
deposited upon the electrode. From the mass measured earlier, the mass
of the Cl would be determined. Alternatively, the Cl gas released could
be collected and weighed, to determine the Cl mass.
7) Water (H2O) takes the shape of triangle, where the two Hydrogen
atoms are bonded to a central Oxygen atom. There is no bond directly
between the Hydrogen atoms. Approximate the energy of a Water
molecule at 300K. Explain each contribution to this energy. What
would the energy of 3 grams of water then be?
To do this, we must determine the degrees of freedom for this molecule.
1) Translational motion in 3D  (3/2)kT
2) Rotation around all three axes(3/2)kT
3) Vibration of each H atom with respect to the O atom
2x2x(1/2)kT=2kT
4) The H atoms can also vibrate with the O-H bonds acting not as a
compression and extension spring, but as a ‘cantilever’. The H atoms
in this case will move closer together and then farther apart as they
vibrate. This mode will add an additional kT
Total average energy would be 7kT.
The energy of 3g of water would be:
 6.022  10 23 molecules 
  2908J
E  7kT * 3g * 
18 g

