13.Calculate the Kinetic Energy of 1 Mole of Nitrogen Molecules at 300 K

13.Calculate the Kinetic Energy of 1 Mole of Nitrogen Molecules at 300 K

12.Which, if either, contains a greater number of molecules, one mole of hydrogen or one mole of oxygen? Which, if either, has more mass?

Answer: One mole of any substance contains Avagadro's number (6.02 x 1023) of particles. Consequently, one mole of hydrogen and one mole of oxygen contain the same number of molecules.

Since they both contain the same number of molecules and the mass of an oxygen molecule is about 16 times that of a hydrogen molecule, one mole of oxygen has about 16 times the mass of one mole of hydrogen.

13.Calculate the kinetic energy of 1 mole of nitrogen molecules at 300 K.


Model Nitrogen as an ideal gas.

We know KE for one particle = 3/2 k T

So for a mole = 3/2 N k T which is 3/2 R T (k = R / NA)

= (3/2) 8.3145 J/(mol K) * 300 K

= 3742 J / mol (or 3.74 kJ/mol)

*Note that at 300 K, gases that behave ideally have the same energy per mol.

14.The molar masses are 349 and 352 for 235UF6 and 238UF6 respectively. The separation of these two isotopes of uranium from a gaseous mixture containing both of them was critically important to US scientists during WWII. How might they have used the theories we have discussed here (think rms) to achieve this task. Draw a sketch of the type of apparatus they may have used.


Their root-mean-square speeds are 149 and 148 m/s respectively. This allowed for enrichment via diffusion separation through a small opening. Due to the small difference in rms speeds, many enrichment stages were necessary and alternative methods were developed.

15.If the speed of each molecule in an ideal gas were tripled, would the temperature also triple?

Answer: No, the temperature will not triple. When the speed of each molecule is tripled then the root-mean-square velocity will also triple. Consequently, the kinetic energy of each molecule will increase by a factor of 9. Therefore, the Kelvin temperature of the gas will also increase by a factor of 9.

16.If the temperature of an ideal gas is doubled from 50°C to 100°C, does the average kinetic energy of the molecules double?

Answer: No, the kinetic energy will not double. In fact, the kinetic energy per particle is related to the absolute temperature through the relationship:

1/2mv2 = (3/2)kT

So, if the Celsius temperature increases from 50°C to 100°C, the absolute temperature (or Kelvin temperature) increases from 323K to 373K. Thus, the fractional increase in kinetic energy is (373/323) ~ 1.15.

17.Can you explain if it is possible for both the pressure and volume of an ideal gas to change without causing the internal energy of the gas to change?

Answer: According to the kinetic theory, the energy of N molecules of an ideal gas at a temperature T (Kelvin!) is (3/2)NkT. Thus, the energy depends only on the temperature. So, if the pressure and volume are changed isothermally there will be no change in internal energy. This means that the changes in pressure (P) and volume (V) have to be such that the product PV remains constant, i.e., if you double the pressure you must halve the volume.

18.If the atoms in a container of helium gas have the same rms speed as the atoms of a container of argon, which one has the higher temperature?

Answer: If the rms speed are the same then the argon atoms have the greater kinetic energy, because the mass of argon atoms is about 10 times that of helium atoms. Since the temperature of an ideal gas is directly proportional to the kinetic energy of the atoms then the temperature of the container of argon atoms (in Kelvins) is about 10 times that of the container of helium atoms.

19.If a slippery cork is pushed into an almost full bottle of wine and released it will slowly slide back out. However, if your pour some wine out and re-insert the cork, the cork does not slide out. Why?

Answer: When the bottle is very full, the volume of air in the bottle above the wine is relatively small. Consequently, when the cork is inserted the reduction in volume that occurs is an appreciable fraction of the original volume. Thus, there is a correspondingly large increase in pressure, which may be sufficient to push the cork out. If some of the wine is removed, the volume of air above the wine is now much larger and so the relative reduction in volume when the cork is pushed in is now much smaller. As a result, the increase in pressure is much smaller and not sufficient to push the cork out.

20.When bubbles rise in a glass of beer, ummmm I mean soda, should they change in size? Should their speed increase, decrease or remain the same as they rise?

Answer: Yes, as bubbles rise from the bottom of a glass their volume increases! Why? Because, as a bubble rises it experiences decreasing pressure. Although we likely haven’t covered this yet, this is due to the fact that there is less ‘weight’ of liquid above it. The ideal gas law tells us that, assuming the beer (I mean Mountain Dew) is at constant temperature, the product of the pressure and volume must be constant also. As a result, the reduction in pressure on the bubble will lead to an increase in volume.

However, the effect of the change in pressure on volume is actually very small in the case of beer (red bull) in a glass. A much larger effect is the increase in volume due to an increase in the number of molecules of gas in the bubble! Why is that? Because a bubble acts as a "nucleation" (or "birth") point for other bubbles; they quickly merge to form a larger bubble. Kinda like raindrops on a window….

But what about the speed of a bubble as it rises? The speed will increase also because the increase in volume leads to an increase in the ‘buoyant force’ on the bubble. The ‘buoyant force’ is equal to the weight of beer (jolt) displaced, which increases as the volume of the bubble increases.


You have recently been promoted to Captain of an Academy Craft for Exploration (ACE) due to excellent performance on earlier scientific inquiry projects. You are on a mission to study the unusual properties of a gas reportedly existing on planet Irona in a far off galaxy in the Leonid cluster.

When you land on the planet, everything seems normal, except the atmosphere seems to be much ‘thicker’ from what you are used to, and you can barely see your hand when you hold it in front of your face.

Quickly you decide to collect a sample and return to the ACE to analyze it. In the lab you subject a sample of the gas to an isothermal pressure test to watch volume response, and you also took a sample and heated it up under constant pressure conditions.

Question 0. Explain in detail the actual apparatus you could have used to accomplish these tests. What I am asking for is the description of the equipment you might use to perform an isothermal pressure test, and also the equipment you could use to heat a gas under constant pressure conditions. Sketch the apparatus also.

On the following page are the test results:

Answer all the questions on the following page.

  1. Using the data seen above, determine a mathematical relationship between Volume and Pressure. Explain if there are any significant differences between how the atmospheric gas on Irona behaves vs. those on earth.
  1. Repeat the previous task, this time focusing on the relationship between Volume and Temperature.
  1. Are there any limitations on Pressure? What are they? Why do you think so?
  1. Can you develop a mathematical expression for the Ideal Gas Law on Irona?
  1. When we derived the ideal gas law, we stated that the Kinetic Energy of gas particles is directly proportional to the temperature. Is this still true on Irona? If not, can you develop a new one?


1. There is a direct relationship as seen from the graph. V = kP. This can be thought

of as V1 / P1 = V2 / P2. The gas is a lot weirder than on earth as it seems to increase in pressure even as its volume expands.

2. There is an inverse relationship, as VT = Constant (hyperbola). This can be thought of as V1 T1 = V2T2. The gas is still weird, as volume drops with increasing temperature.

3. The only pressure limitation is that it could never be zero, or the volume would be zero.

4. PT = nRV (remember R would now have different units, what would they be?)

5. Well I guess it is true that KE depends on T, it’s just that KE = k / T now. So I’d say that KE is INVERSELY proportional to T.