In Theory, We Trust - Kinetic Theory Solutions

A bonus assignment for tonight - if you check your answers, please log in and write a comment (go to the bottom of the post for the link) stating the name of your favorite ideal gas. If you don't have one, choose Helium.

50. The gas starts at 0 degrees C, or 273 K. Since v = (3RT/m)^1/2, in order to double v, the temperature must be multiplied by 4. Thus the temperature must be 1092 K.

52. Following the similar logic as with #50, If the pressure doubles, and volume stays the same, the ideal gas equation says that the temperature will also double. This will make the v_rms increase by a factor of the square root of 2.

55. You could get to the solution of this equation by looking at the derivation we did today. We ended up with the fact that P = (1/3)m*N*v^3/(V).

The quantity m*N is the total mass of the gas contained in the container, as it's the mass of a single molecule times the number of molecules. The quantity M*N/V must then bethe density of the gas.

Also from the derivation, the v in our expression is the mean square speed of the molecules in the gas.

Thus P = (1/3)*density*v^2

There are other ways to derive it using the other formulas we came up with today, so if your method is different, that is fine.

80. Since you know the temperature, you can directly calculate the v_RMS = 183.5 m/s.

Using the result of question 5, we have a relationship between the density, v_RMS, and the pressure.

The density will be one atom/cm^3 = 1 amu * 1.66 x 10^-27 kg*(10⁶cm^3/m^3) = 1.66 x 10^-21 kg/m^3.

Substituting, the pressure is 1.9 x ^-17 Pa = 1.84 x ^-22 atm

81. When 10 meters below the surface, the absolute pressure is Patm + rho*g*h = 199,300 Pa.
At the surface, the pressure is 101,300 Pa. You also know the initial volume underneath the water, as well as the fact that the moles of air in the diver's lungs are constant as he/she rises to the surface.

If we assume the rise happens, quickly, we can also say that the temperature is approximately constant.

Thus P1V1/P2V2 = nRT1/nRT2. The entire right side of the equation divides out to 1.

Thus V2 = P1V1/P2 = 10.8 L. I personally would NOT want my lungs stretched to three times their original volume, nor would I expect it to be medically advisable for ANYONE to have this done to them.

This is one of the reasons why it's important for scuba divers to take their time rising to the surface after a dive. One of them is that it causes a condition called 'the bends', where dissolved gas comes out of solution in the blood stream. This is a bad thing. Another is that, without releasing any gas in the lungs during the resurfacing process, the lungs will expand rapidly during the rise, possibly causing tearing in the tissue.