Week 11: Ideal Gases & Kinetic Theory

\(\Delta\) ← if you see the word "Delta" instead of a triangle, please refresh the page! Thanks :)

[1] Overview⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

• Using the ideal gas law

  • \(PV=nRT\)
  • remembering to use Kelvin

• Special gas-process cases

  • constant pressure
  • constant volume
  • combined gas-law reasoning

• Relating moles, molecules, and molar mass

  • using Avogadro's number
  • converting between molecular mass and molar mass

• Kinetic theory of gases

  • why average velocity can be zero
  • why temperature is tied to average kinetic energy
  • why lighter molecules move faster at the same temperature

• Average translational kinetic energy, rms speed, and internal energy

  • \(\overline{K} = \tfrac32 k_B T\)
  • \(v_{\mathrm{rms}} = \sqrt{3RT/M}\)
  • \(U = \tfrac32 nRT\) for a monatomic ideal gas

• Gas problems involving fluids and forces

  • using \(P = P_{\mathrm{atm}} + \rho gh\)
  • pressure differences
  • spring forces
  • force/area ideas together with the ideal gas law

[2] Practice Problems⚓︎

If you're comfortable with the "★★★" problems, you should do great during recitation.

If you want to print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely.

Difficulty key:

• ★☆☆ = beginner
• ★★☆ = standard
• ★★★ = challenging / multi-step

[2.1] Ideal Gas Law and Gas Processes⚓︎

1. ★☆☆
   A gas is held in a vertical cylinder with a frictionless movable piston, so the pressure stays constant. The gas column height increases from \(16.0 \ \mathrm{cm}\) to \(28.0 \ \mathrm{cm}\) while the temperature rises from \(18.0^\circ \mathrm{C}\). What is the final temperature of the gas?

   Solution

   At constant pressure, an ideal gas obeys

   \[\frac{V_1}{T_1}=\frac{V_2}{T_2}.\]

   Since the cylinder has constant cross-sectional area, volume is proportional to height, so

   \[\frac{h_1}{T_1}=\frac{h_2}{T_2}.\]

   Convert the initial temperature to Kelvin:

   \[T_1 = 18.0 + 273 = 291 \ \mathrm{K}.\]

   Then

   \[T_2 = T_1\left(\frac{h_2}{h_1}\right)=291\left(\frac{28.0}{16.0}\right)=5.09 \times 10^2 \ \mathrm{K}.\]

   So the final temperature is about

   \[T_2 \approx 510 \ \mathrm{K}.\]

   In Celsius,

   \[T_2 \approx 510 - 273 = 237^\circ \mathrm{C}.\]

   
   Answer: \(T_2 \approx 5.10 \times 10^2 \ \mathrm{K}\), or about \(237^\circ \mathrm{C}\).

2. ★★☆
   A rigid \(3.60 \ \mathrm{L}\) tank contains an ideal gas at \(1.35 \times 10^5 \ \mathrm{Pa}\) and \(320 \ \mathrm{K}\). The gas is cooled to \(260 \ \mathrm{K}\).

   • (a) Find the final pressure.
   • (b) Find the number of molecules in the tank.
   Solution

   Since the tank is rigid, \(V\) is constant, so

   \[\frac{P_1}{T_1}=\frac{P_2}{T_2}.\]

   Therefore,

   \[P_2 = P_1\left(\frac{T_2}{T_1}\right) = (1.35 \times 10^5)\left(\frac{260}{320}\right)=1.10 \times 10^5 \ \mathrm{Pa}.\]

   For the number of molecules, first find the number of moles from the initial state:

   \[n = \frac{PV}{RT}.\]

   Convert volume:

   \[3.60 \ \mathrm{L} = 3.60 \times 10^{-3} \ \mathrm{m^3}.\]

   Then

   \[n = \frac{(1.35 \times 10^5)(3.60 \times 10^{-3})}{(8.31)(320)}=0.183 \ \mathrm{mol}.\]

   The number of molecules is

   \[N = nN_A = (0.183)(6.02 \times 10^{23})=1.10 \times 10^{23} \text{ molecules}.\]

   
   Answer: (a) \(1.10 \times 10^5 \ \mathrm{Pa}\); (b) \(1.10 \times 10^{23}\) molecules.

3. ★★★
   A weather balloon has a volume of \(2.20 \ \mathrm{m^3}\) when the pressure is \(0.95 \ \mathrm{atm}\) and the temperature is \(25.0^\circ \mathrm{C}\). It rises to a region where the pressure is \(0.42 \ \mathrm{atm}\) and the temperature is \(-18.0^\circ \mathrm{C}\). Assuming the amount of gas stays constant, what is its new volume?

   Solution

   Use the combined gas law:

   \[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}.\]

   Convert temperatures to Kelvin:

   \[T_1 = 25.0 + 273 = 298 \ \mathrm{K}, \qquad T_2 = -18.0 + 273 = 255 \ \mathrm{K}.\]

   Solve for \(V_2\):

   \[V_2 = \frac{P_1V_1T_2}{P_2T_1}.\]

   Plug in the values:

   \[V_2 = \frac{(0.95)(2.20)(255)}{(0.42)(298)} = 4.26 \ \mathrm{m^3}.\]

   
   Answer: \(V_2 \approx 4.26 \ \mathrm{m^3}\).

[2.2] Kinetic Theory, rms Speed, and Internal Energy⚓︎

1. ★☆☆
   In a gas at thermal equilibrium, the average speed of the molecules is not zero, but the average velocity is zero. Explain why.

   Solution

   Speed is a scalar, so it only tells you "how fast." Velocity is a vector, so direction matters.

   In a gas at equilibrium, molecules move in all directions. On average, there are just as many molecules moving in the \(+x\) direction as in the \(-x\) direction, and similarly for the other directions. Because those vector contributions cancel out, the average velocity is zero.

   But each molecule is still moving, so the average speed is not zero.

   
   Answer: The average velocity is zero because the velocity vectors cancel in all directions, while the average speed is nonzero because molecules are still moving.

2. ★★☆
   A sample of neon gas is at \(350 \ \mathrm{K}\).

   • (a) Find the average translational kinetic energy per atom.
   • (b) Find the rms speed of the neon atoms.
   • Use \(M_{\mathrm{Ne}} = 20.2 \times 10^{-3} \ \mathrm{kg/mol}\).
   Solution

   For any ideal gas,

   \[\overline{K} = \frac32 k_B T.\]

   So

   \[\overline{K} = \frac32 (1.38 \times 10^{-23})(350)=7.25 \times 10^{-21} \ \mathrm{J}.\]

   For rms speed,

   \[v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}.\]

   Therefore,

   \[v_{\mathrm{rms}} = \sqrt{\frac{3(8.31)(350)}{20.2 \times 10^{-3}}}=6.57 \times 10^2 \ \mathrm{m/s}.\]

   
   Answer: (a) \(7.25 \times 10^{-21} \ \mathrm{J}\) per atom; (b) \(6.57 \times 10^2 \ \mathrm{m/s}\).

3. ★★★
   A \(3.50 \ \mathrm{mol}\) sample of a monatomic ideal gas is heated from \(280 \ \mathrm{K}\) to \(420 \ \mathrm{K}\).

   • (a) Find the change in internal energy.
   • (b) By what factor does the rms speed change?
   Solution

   For a monatomic ideal gas,

   \[U = \frac32 nRT,\]

   so the change in internal energy is

   \[\Delta U = \frac32 nR\Delta T.\]

   Here,

   \[\Delta T = 420 - 280 = 140 \ \mathrm{K}.\]

   Thus,

   \[\Delta U = \frac32 (3.50)(8.31)(140)=6.11 \times 10^3 \ \mathrm{J}.\]

   For the rms speed,

   \[v_{\mathrm{rms}} \propto \sqrt{T}.\]

   So the factor change is

   \[\frac{v_{\mathrm{rms},2}}{v_{\mathrm{rms},1}}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{420}{280}}=\sqrt{1.50}=1.22.\]

   
   Answer: (a) \(\Delta U = 6.11 \times 10^3 \ \mathrm{J}\); (b) the rms speed increases by a factor of \(1.22\).

[2.3] Gas Pressure with Fluids, Springs, and Forces⚓︎

1. ★★☆
   A small air bubble has volume \(42.0 \ \mathrm{cm^3}\) at the surface of a freshwater lake, where the pressure is atmospheric. It is taken down to a depth of \(5.50 \ \mathrm{m}\). Assume the temperature stays constant.

   • (a) Find the pressure at that depth.
   • (b) Find the bubble's volume at that depth.
   Solution

   The pressure at depth \(h\) in water is

   \[P_2 = P_{\mathrm{atm}} + \rho gh.\]

   So

   \[P_2 = 1.013 \times 10^5 + (1000)(9.80)(5.50)=1.55 \times 10^5 \ \mathrm{Pa}.\]

   Since the temperature is constant, Boyle's law applies:

   \[P_1V_1 = P_2V_2.\]

   Therefore,

   \[V_2 = \frac{P_1}{P_2}V_1 = \frac{1.013 \times 10^5}{1.55 \times 10^5}(42.0)=27.4 \ \mathrm{cm^3}.\]

   
   Answer: (a) \(1.55 \times 10^5 \ \mathrm{Pa}\); (b) \(27.4 \ \mathrm{cm^3}\).

2. ★★☆
   An ideal gas is confined in a vertical cylinder by a massless piston attached to an ideal spring. There is a vacuum outside the cylinder. The piston has cross-sectional area \(A = 2.50 \times 10^{-3} \ \mathrm{m^2}\). Initially, the gas volume is \(5.20 \times 10^{-4} \ \mathrm{m^3}\), the gas temperature is \(295 \ \mathrm{K}\), and the spring is stretched by \(6.00 \ \mathrm{cm}\). The gas is heated until the spring stretch becomes \(8.50 \ \mathrm{cm}\).

   • (a) Find the final volume.
   • (b) Find the final temperature.
   Solution

   Because the outside is a vacuum, the gas pressure balances only the spring force:

   \[PA = kx \qquad \Rightarrow \qquad P \propto x.\]

   So

   \[\frac{P_2}{P_1} = \frac{x_2}{x_1} = \frac{8.50}{6.00} = 1.42.\]

   The volume change comes from the piston moving a distance \(\Delta x\):

   \[V_2 = V_1 + A\Delta x = V_1 + A(x_2-x_1).\]

   Convert the stretches to meters:

   \[x_1 = 0.0600 \ \mathrm{m}, \qquad x_2 = 0.0850 \ \mathrm{m}.\]

   Then

   \[V_2 = 5.20 \times 10^{-4} + (2.50 \times 10^{-3})(0.0850 - 0.0600)\]

   \[V_2 = 5.83 \times 10^{-4} \ \mathrm{m^3}.\]

   Now use the ideal-gas relation with fixed \(n\):

   \[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}.\]

   So

   \[T_2 = T_1\left(\frac{P_2}{P_1}\right)\left(\frac{V_2}{V_1}\right).\]

   Therefore,

   \[T_2 = 295\left(\frac{8.50}{6.00}\right)\left(\frac{5.83 \times 10^{-4}}{5.20 \times 10^{-4}}\right)=4.68 \times 10^2 \ \mathrm{K}.\]

   
   Answer: (a) \(V_2 = 5.83 \times 10^{-4} \ \mathrm{m^3}\); (b) \(T_2 \approx 468 \ \mathrm{K}\).

3. ★★★
   A rigid glass bottle is sealed with a cork of cross-sectional area \(2.20 \ \mathrm{cm^2}\). Initially, the air inside is at \(24.0^\circ \mathrm{C}\) and atmospheric pressure. The cork will start to slide out when the upward force from the gas exceeds the downward atmospheric force by \(8.00 \ \mathrm{N}\). Assuming the bottle's volume stays constant, what minimum temperature must the air inside reach before the cork starts to move?

   Solution

   The cork starts to move when the net upward force from the pressure difference reaches \(8.00 \ \mathrm{N}\):

   \[\left(P_{\mathrm{in}}-P_{\mathrm{atm}}\right)A = 8.00.\]

   So the required inside pressure is

   \[P_2 = P_{\mathrm{atm}} + \frac{8.00}{A}.\]

   Convert the area:

   \[2.20 \ \mathrm{cm^2} = 2.20 \times 10^{-4} \ \mathrm{m^2}.\]

   Then

   \[P_2 = 1.013 \times 10^5 + \frac{8.00}{2.20 \times 10^{-4}} = 1.38 \times 10^5 \ \mathrm{Pa}.\]

   Since the bottle is rigid, \(V\) is constant, so

   \[\frac{P_1}{T_1}=\frac{P_2}{T_2}.\]

   Convert the initial temperature to Kelvin:

   \[T_1 = 24.0 + 273 = 297 \ \mathrm{K}.\]

   Solve for \(T_2\):

   \[T_2 = T_1\left(\frac{P_2}{P_1}\right)=297\left(\frac{1.38 \times 10^5}{1.013 \times 10^5}\right)=4.04 \times 10^2 \ \mathrm{K}.\]

   Converting back to Celsius,

   \[T_2 \approx 404 - 273 = 131^\circ \mathrm{C}.\]

   
   Answer: The air must reach about \(4.04 \times 10^2 \ \mathrm{K}\), or about \(131^\circ \mathrm{C}\).