Cindy Spangler
February 12, 2007
Lab summary of KMT and equations of state
In the previous labs, we investigated equations used to describe gases; however these equations make many assumptions that limit the ability to accurately describe the behaviors of the gas particles. The assumptions treat all gas particles the same even though there are several differences in each gas. For example, gas particles are assumed to have no volume. So when looking at two gaseous atoms, Argon and Helium, Helium will display a more ideal gas behavior because the same number of atoms of He will occupy less empty volume than Argon.
We also see this with attractive forces between the gas particles. For gaseous water and oxygen, oxygen would act more like an ideal gas because the molecules of oxygen do not have the hydrogen bonding that water molecules have. But then this raises the question of what factor is more important in affecting the behavior related to an ideal gas: particle size or attractive forces? The size of a water molecule may be smaller (therefore occupying less empty volume) than the oxygen, but has attractive forces between the water molecules. Attractive forces may also cause a change in the expected pressure exerted by the gas particles. Gas particles exert a force on the walls of a container because they have mass; if the mass is not recognized because of the limits of KMT, then the pressure observed will be different than calculated.
Another approximation that KMT makes is that the velocity components are independent in all directions and the combined probability is given by
F ( v )
 f ( v x
) f ( v y
) f ( v z
) eq. 1 where this is the probability of the velocity of the particle. Taking the natural logarithm of each side of equation
1, we find
 ln F
 v x
( v )

 ln f
 v x
( v x
) eq. 2
Assuming that v
 v
2 x
 v
2 y
 v
2 z
, we found

( v )
 v x
 y , z
 v x v
Eq. 3
Combining both equations, we are able to find the probability of v as a function of v , x
1 d ln F ( v )

1 d ln f ( v x
) dv x
Eq. 4 v dv v x where the left hand side of equation 4 is a constant because it does not vary with the x, y, or z component of velocity. We can then determine the probability as a function of velocity because we know that the left hand side of the equation is constant. The resulting equation is the probability of particles at the velocity
 av
2
F ( v )

Ae 2 dv . eq. 5
We now want to relate the volume and the velocity of the gas particles together. Knowing the fraction of particles in at a volume is given by

F ( v )

1
N dN ( v ) dv
Eq. 6
And that the change in volume of a sphere is dV

4
 v
2 dv Eq. 7 where v is the radial component and velocity of the particle. Combining equations 5 and 6 together by setting
F( v ) equal for each equation we get
1
N dN ( v )
 dv
Ae
 av
2
2 4
 v
2 dv eq. 8
We then integrate equation 8 to find that A
 a
2

3 / 2 and substitute a
 m k
B
T
to find that
F ( v )


 m
2
 k
B
T


3 / 2 e
 m ( v
2 x
 v y
2
2 k
B
T
 v z
2
) eq. 9
These equations are relating the number of the particles to specific velocities, providing what is known as the
Maxwell-Boltzmann distribution, where m is the mass of the particle, v is the speed of the molecule, k is the
Boltzmann constant, and T is the absolute temperature of the system. To find the velocity for a 3D model, we will look at v 4 . From this relationship, we find that the velocity increases as the dimensions of the box increases. But why is this?
From the maple worksheet, we are looking at equation 9 in more detail. We know from physics laws that the average kinetic energy of the particle is given by the kinetic energy of motion and also relate it to the dimensions of the container

E kinetic

1
2 mu
2 
1
2
DkT where u is the average velocity of the particle squared (to be squared again) and D is the dimensions of the container. In the model that we are looking at in Maple, we are looking at He gas. We will now graph the distribution for the probability of finding the particles in the range of [ v, v+dv ] in a 2D box. eq. 10
Where the T= 100 K, 300 K, and 1000 K for the red, blue and green distribution lines respectively. At the temperature increases, the probability of finding gas particles at faster speeds increases and the average speed

increases for the system. With the increase in temperature, the deviation from the average speed increases meaning that you find more particles at different speeds than are observed at lower temperatures. From the plot, the average speed would be larger than the most probable particle speed because the distribution of the particles spread out significantly and have a long tail where more of the particles are at faster velocities. We can also see that the distribution is affected by the mass of the particle.
This is the distribution for a particle with half the mass of the previous plot. The distribution is spread out more for lighter particles and the width increases for the most probable speed and the average speed. Therefore as mass increases, the average particle speed and most probable speed decreases.
IDEAL GAS SIMULATION
We find that gases can be modeled as well with random number generation as with Newton’s Law of motion. So we use maple and random number generation to model an ideal gas. To understand how maple works, I placed in new seed numbers to see whether male was a random number generator. If the same seed is used, the same numbers are given from maple. We used the original seed to plot the random numbers generated. The resulting plot was created
Five hundred points are represented in the graph above and model the locations of gas particles. There appears to be no patterns from this sequence of random numbers and the circles cover most of the plot. This is the most probable distribution for the particles to cover the entire plot. If the plot was to be divided, I would expect to count the number of particles (for this model 500) divided by the number of equally spaced intervals (10 for the example). This would mean that 50 particles would be in each of the intervals as the most probable distribution.
We can see this by looking at a histogram of the random numbers generated.

This is a histogram for 500 particles where the y-axis is the ratio of one cell to the next. The x-axis represents the number of equal divisions in the larger container. The cells have an approximate one to one ratio, but this ratio is more evident as the number of particles is increased as shown for the following histogram for 1500 particles.
The ratio is more uniform over the ten cells; the deviation from the one to one ratio decreases as the number of particles increases. The histogram would follow a uniform distribution when all the ratios were one to one, meaning the same number of particles is found in each cell. From the histogram it is easier to see that the random number generation does provide a uniform distribution over all the bins and will effectively model an ideal gas.
If we have non-uniform random numbers, we have to truncate the velocities to a finite value. This is done by using approximate values that would be appropriate for a given temperature. This is reasonable because according to the Boltzmann distribution of particles at a certain temperature, most of the particles are grouped together at a most probable speed and if the maximum velocity is greater than this value, the truncation occurs where 90 to 95 percent of the particles are accounted for in the distribution. After the velocities of the particles are truncated, we have a histogram that is in the similar shape as the Boltzmann distribution shown in the first plots. As the number of particles increases, the histogram plot became smoother with fewer deviations from the plotted Boltzmann distributions done in the beginning of the lab. When temperature was increased, the distribution spreads out and becomes more symmetric, similar to what we observed in the beginning. When comparing the graphs from the histogram and the actual distribution function, the graphs are very similar because

they are plotting the same thing and have approximations that are made with the velocities of the particles. They vary in the same manner when variables are changed. Differences in the analytical solution and the calculated average velocity differ because in the calculated value, we made velocity approximations to decrease the amount of time spent calculating the values. This results in the average velocity being lower than the actual value. To make the calculations more accurate, we can increase the velocity and the number of particles, but this takes more time to complete all the calculations.
We now look at the motion of the particles using maple and find the pressure. We find that using maple, there are differences in the simulation value and the analytical solution because we have a limited velocity and a limited number of particles. This limit causes the simulation value to be lower than the analytical solution. If a different seed is used to determine the pressure, we find that the values for the pressure remain constant and the graphical trends are the same. Even though a new set of random numbers are used, the approximations used in the equations work for all random number generation.
We played with the values of volume and the number of particles and found that as N increases, the pressure increased because the volume and number of particles are dependent on area. If the temperature is changed, the pressure of the system changes. The following graph provides the data where temperature and pressure are graphed.
Pressure v. Temperature
8.60E-20
8.40E-20
8.20E-20
8.00E-20
7.80E-20
7.60E-20
7.40E-20
7.20E-20
P
0 200 400 600 800 1000 1200
Temperature (K)
We see that the pressure and temperature are directly proportional at higher temperatures. This is consistent with the ideal gas law, PV=nRT.
The reason that at lower temperatures the relationship does not seem to hold is because of the limits that we are using in order to determine the pressure. We have a limited number of particles and velocity is truncated. The pressure decreased when the volume of the box was increased and the following graph shows how pressure changed with temperature.
Pressure v. Temperature for L=0.5
4.00E-21
3.50E-21
3.00E-21
2.50E-21
2.00E-21
1.50E-21
1.00E-21
5.00E-22
0.00E+00
P
0 200 400 600 800 1000 1200
Temperature (K)
As the volume increased, the graph is less consistent with the graph obtained above as a result from the limits we placed on the calculations. This does not look like it models the ideal gas law. We then looked at pressure and the affect of the number of particles in the system.

Pressure v. number of particles
1.80E-19
1.60E-19
1.40E-19
1.20E-19
1.00E-19
8.00E-20
6.00E-20
4.00E-20
2.00E-20
0.00E+00
P
0 50 100 150 200 250
N
We found that as the number of particles increased in a container held at a constant temperature and volume, the pressure increased linearly. This is what was expected using the equation for the ideal gas law. However this is only an approximation and has limited number of particles, there is particle interaction, and the particles take up space.
For the second model we worked with how a real gas varies from an ideal gas model. Using the ideal gas law,
PV=nRT, we solve for internal pressure of the gas. For the ideal gas law, the type of gaseous molecule does not cause variation in the PV calculation. PV is equal to a constant when the number of moles of gas and the temperature are held constant. As pressure increases, the relationship to PV is horizontal and constant.
Looking at the ideal gas law closer, we find that compressibility factor, Z, of a gas. Compressibility is given by
Z

P V
,
RT
where V is the molar volume,
V n eq. 11
For an ideal gas, Z=1 (with no units) and the compressibility is not dependent on pressure of the system.
Knowing this, we are able to expand the ideal gas law to account for the space occupied by the gas molecules.
Because the unoccupied volume is now less than with the ideal gas law, the pressure will be greater for when the particles have volume. To correct for the space being occupied by the particles, we use equation 12:
P
 nRT
( V
 nb )
eq. 12
According to this model of a gas, a real gas will have a greater pressure than an ideal gas and the compressibility factor will always be greater than one. For this model of a gas, compressibility is expressed by
Z

1

Pb
RT
eq. 13
Where the second term on the right hand side will never be negative. From this equation and using methane as an example (b=0.0428 L mol -1 ) we see how Z varies with pressure.

Z vs. P for methane
3.5
3
2.5
2
1.5
1
0.5
0
-200 0 200 400 600 800 1000 1200
Pressure (bar)
We see that Z increases as pressure increases. The molar volume of a gas is greater than the molar volume of an ideal gas because the compressibility factor is larger than for the ideal gas. As Z increases above one, the gas becomes less compressible than an ideal gas.
In KMT of gases, it assumes that there are no attractive forces between the gases, which is an invalid assumption for a real gas; gas particles will have attraction and repulsion between the molecules. If the gaseous particles were attracted to each other, the pressure of the gas would decrease because the molecules will remain closer to each other and have fewer collisions with the walls. The opposite would be true for molecules that were repulsed. To modify our equation 13, we want to subtract the attractive nature of the molecules. This provides us with equation 14
P

V nRT
 nb
 an
V
2
2 eq. 14
The second term on the right hand side accounts for Van der Waal’s attractive forces on pressure. For a molecule that has partial charges, van der Waal’s would be greater than a non-polar molecule. From the attractive forces due to van der Waal’s forces, the compressibility factor, Z, can be less than one. This is shown by the following graph
Z vs. P
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-100 0 100 200 300 400 500 600
Pressure (bar)
This shows that because of the attractive forces of methane, the gas is able to be compressed more than an ideal gas. At low pressures, van der Waal’s is the factor causing the deviations from the ideal gas law, but at high pressure, the volume of the molecules cause the greatest deviation from ideality.
Changing the temperature of the equation causes the graph to vary.

Z vs. Pressure
2.5
2
1.5
1
0.5
25c
100c
300
Pressure (bar)
0
0 500 1000 1500 2000
From the graph, we can see that at low pressure and high temperature, the gas behaves more ideally than at low temperatures and high pressure; this means that the ideal gas law models these situations better. We also know that at high temperatures and low pressures, the gas must have the same limits that are assumed with KMT; lower attractive/repulsion factor and the volume of the molecules doesn’t affect the volume of the container. We see that as the pressure approaches zero, Z approaches one for all gases. All variation from Z=1 can be thought of as a deviation from an ideal gas.
The slope of Z would increase if the volume of the molecule is greater. The larger the non-interactive particles, the greater the slope of Z with respect to pressure.
We want to be able to calculate the molar volume of the gas using Newton method. We first write the van der
Waals gas equation as a cubic polynomial, shown in equation 15.
P V
3 
V
2
( Pb

RT )
 a V
 ab

0 eq. 15
By entering this equation into excel and using the relationship x x

1
 x n
 f ( x n
) f ' ( x n
) eq. 16
We use the ideal gas law for the first approximation. From Newton’s approximation method, we get 0.09604
L/mol for ethane. This is compared to 0.071 L/mol; which is 35% more than the experimentally determined value. This means that the limitations of the van der Waals gas equation and of Newton’s method do not approximate the molar volume of ethane within experimental uncertainty. By changing the equation we are able to obtain better approximations; however the formulas become more complicated. Using the Redlich-Kwong equation
P

V
RT

B

T
A
1 / 2
V ( V

B )
eq. 17
A and B are parameters that depend on the identity of the gas. We can also find molar volume as a cubic polynomial. We find f ( x n
)

V
3 
V
2
(
RT
P
1 / 2
)

V ( B
2 
BRT
1 / 2

A
PT
1 / 2
)

AB

0 eq. 18
P PT
1 / 2
Retyping this was almost as fun as deriving it; but the complexity of the equation provides a better approximation of ethane. Using Excel, the approximation is 7% different from the experimental value and within experimental

uncertainty. This is a better way to calculate the molar volume, but as accuracy increases, the complexity of the equation does, which means that there is a trade off in time and accuracy.
Looking at the intermolecular potential in the hard sphere model, the particles have completely elastic collisions, therefore the only moment that they have potential is at the time that the two particles are in contact. There is infinite potential at the moment that the two particles are at R from each other and at all other distances, there is no potential. This model is similar to model 3, Real gases/size factor because the radius and volume are the only factors affecting the outcome of the model. There are no attractive forces considered in either model. Using the hard sphere model, we derive the virial coefficient to be
B
2 v
( T )

2

N
A
[ r
3
3
] eq. 17
This means that the compressibility factor is greater than one and that the attractive forces are the most important factors for causing the deviation from ideality. If attractive forces are used in the model, we get a different graph.
The graph has infinite potential and negative potential depending on the radius of the molecule. The graph is dependent on the attractive forces, the magnitude of attractive forces and the distance to the center. Expanding this equation, we are able to find the sum of the potential for gases that do not have completely inelastic collisions. We find the virial values for Ar and carbon dioxide:
T=273K
T=373K
Argon
Carbon
Dioxide
-0.007847345 -0.061546507
0.003780865 -0.030383421
When we used the virial for the hard sphere model of Argon, we see that they differ by a factor of ten. The hard sphere model is ten times greater than the square well potential, which accounts for the attractive forces. At the same temperature, both particles vary because of the interactions that they have with each other. Argon is less attracted to other Argon values; therefore a more positive virial value.