Reading Quiz

Question 1:

In the microscopic model for an ideal gas discussed in the reading, what is the physical origin of the pressure exerted by the gas?

Answer:

In this model, gas particles are constantly colliding with the walls of the container. These collisions exert a force per unit area, and thus result in pressure.
  1. The physical origin of the pressure exerted by the gas is the elastic collisions of the gas particle off of the walls of the chamber and the piston.
  2. Gas molecules colliding with the piston at high velocities.
  3. The pressure exerted by the gas comes from the force exerted when the molecule(s) collide(s) with the walls of the container.
  4. The pressure comes from the gas molecule(s) bouncing off the walls of the container, thus transferring energy through physical contact to the walls.
  5. The pressure exerted by the gas is the result collisions between the gas molecules and the walls of the container.
  6. Pressure is really just how often the molecule slamms into the piston or a wall. The more often a molecule hits the wall the higher the pressure.
  7. The pressure is forces exerted from collisions of molecules on an area. We can average the forces and frequency of collisions of many particles to understand pressure's dependancies.
  8. kinetic energy of a single gas molecule undergoing elastic collisions in a cylinder.
  9. "periodically the molecule crashes into the piston and bounces off, exerting a relatively large force on the piston for a brief moment."
  10. The physical origin of the pressure exerted by the gas is the force per unit area exerted on the piston and the other walls of the cylinder.

Question 2:

The reading claims that small molecules moving around at ordinary temperatures are bouncing around at hundreds of meters per second. Verify this claim by determining the rms speed of a nitrogen (N2) molecule at room temperature (this is Problem 1.18).

Answer:

Using eq. (1.21) and that the molar mass of N2 is 28 g/mol, we find that the rms speed of nitrogen is 517 m/s.
  1. Using the atomic mass of a nitrogen molecule as 28 amu, v_rms = 520 m/s
  2. v_rms=sqrt((3kT)/m) T= 300 K m= 2 * (14 g/mol / 6.022*10^23 atoms/mol)= 4.64 * 10^-23 g = 4.64 * 10^-26 kg k= 1.38 * 10^-23 J/K v_rms= 517 m/s
  3. From (1.21), v_rms = sqrt(3kT/m), and at room temperature kT = 4.14e-21 J (1.18) m_N2 = 2*14*1.661e-27 kg v_rms = sqrt(3*4.14e-21/(2*14*1.661e-27)) v_rms = 516.8 m/s
  4. rms speed of a nitrogen molecule at room temperature comes to be about 516 m/s
  5. I get about 730 m/s
  6. Vrms = 520 m/s (using eq 1.21 and mass of N2 to be 4.6 x 10^-26 kg)
  7. v_rms (N2 @ 300K) = Sqrt[3kT/m] = Sqrt[(3 * 4.14 x 10^-21 J)/(4.6 x 10^-26 kg (* by Google *) )] = 436 m/s
  8. I found the velocity to be 543 m/sec.
  9. Rougly 500 m/s
  10. 1.216x10^-19 m/s is my calculated vrms. My number doesn't agree with the suggested velocity.

Question 3:

Does a system consisting of diatomic molecules at room temperature have more or less thermal energy than a system consisting of the same number of monatomic molecules, also at the same temperature? Briefly explain your reasoning.

Answer:

Diatomic gas has more thermal energy than monatomic gas. At room temperature, a diatomic molecule has two rotational degrees of freedom in addition to its three translational degrees of freedom. A monatomic gas only has three degrees of freedom. So using eq. (1.23), the diatomic gas more thermal energy. Here we assumed that the vibrational degrees of freedom are "frozen out" at room temperature, which is reasonable.
  1. At the same temperature and with the same number of molecules, the diatomic molecules have more thermal energy at room temperature because they have more degrees of freedom. A monotomic molecule can only move with translational motion, while a diatomic molecule can also rotate about two axes. Using the total thermal energy formula: U_thermal = N x f x (1/2)kT, the diatomic molecules have a bigger f, thus have more thermal energy.
  2. The diatomic molecules would have more energy because they have more degrees of freedom.
  3. From equation (1.23), U_thermal = N*f*1/2*kT is proportional to the number of degrees of freedom. Thus a system of diatomic molecules has more thermal energy, since it has more degrees of freedom.
  4. The system of diatomic molecules should have more thermal energy due to the equipartition theorem because they have more degrees of freedom, due to the shape of the molecules and their ability to rotate in various directions.
  5. The diatomic system has more thermal energy because there are more degrees of freedom and the total thermal energy is given by equation 1.23 (U = N f kT/2). Both systems have three translational degrees of freedom; however, the diatomic molecules have two rotational degrees of freedom.
  6. The system containing diatomic molecules have more degrees of freedom which according to eq. 1.23 results in a higher thermal energy. The degrees of freedom in the diatomic system result from rotations and form vibrations (which don't always contribute at room temperature because of QM).
  7. In general, at RT a system of diatomic molecules has more thermal energy than a system of monatomic molecules because there are more degrees of freedom for motion.
  8. since diatomic molecules have more degrees of motional freedom they must have more energy than monoatomic molecules. Diatomic molecules can also convert their rotational energy to translational energy.
  9. Diatomic molecules have more degrees of freedom that contribute to the thermal energy, namely rotational and vibrational, therefore it should have more thermal energy.
  10. Diatomic molecules have the potential to use vibrations for thermal energy so you would think that diatomic molecules would have more thermal energy at room temperature than monoatomic molecules at the same temperature, but at room temperature this vibrational kinetic energy is frozen out meaning that the diatomic molecules will have the same thermal energy as monoatomic molecules at room temperature.

Question 4:

What (if any) were the conceptual or mathematical difficulties you had with the reading?

Answer:

Your responses below.
  1. I understand that there are quantum mechanical reasons why diatomic molecules cannot rotate along the axis where the molecules lie, but is it explained anywhere in this book? With the bedspring model of the crystalline solid, is it possible within the theory to take one of the molecules and twist it, then release it? If it were an actual set of springs and balls, this would be possible.
  2. The degrees of freedom seem to be based on cartesian coordinates- one degree of freedom for x, y, and z. How would they work in, say, polar coordinates?
  3. I don't feel like I have a clear idea what a degree of freedom is.
  4. Nothing too bad, I'm surprised at how simple this whole 'degrees of freedom' thing appears to be after all the confusion it has caused in past courses, not to say that it is a simple concept, of course...
  5. None. The step-by-step mathematical derivations are very helpful
  8. None as of yet.
  9. N/A
  10. I had no difficulties with the reading for today. My work on problem 1.18 is flawed though so I guess I don't understand that calculation.