Unit 13: Thermal Physics — Short Questions

Yes. The average translational kinetic energy of a gas molecule is <K.E.> = (3/2)kBT, which depends only on the absolute temperature T, not on the mass or identity of the molecule. For example, at room temperature (300 K) both oxygen (O₂) and nitrogen (N₂) molecules in air have the same average kinetic energy, even though their rms speeds differ because of their different masses (nitrogen, being lighter, moves faster than oxygen).

Yes. Since both gases in the vessel are in thermal equilibrium at the same temperature T, and <K.E.> = (3/2)kBT depends only on T, the mean kinetic energy per molecule is the same for both gases. However, since the gases have different molecular masses, their rms speeds are different (the lighter gas has the higher rms speed) even though their average kinetic energies are equal.

From P = (1/3)ρ<v²>, pressure is directly proportional to density when the rms speed (i.e. temperature) is unchanged. Therefore, doubling the density of the gas doubles its pressure.

Heating the gas increases its temperature, which increases the average kinetic energy and hence the rms speed of the molecules, since <K.E.> ∝ T. The faster-moving molecules strike the container walls more frequently and with greater momentum change per collision, so the force exerted per unit area on the walls increases — i.e. the pressure of the gas increases.

The rms speed v(rms) = √(3kBT/m) depends only on the temperature and the mass of a single molecule, not on the number of molecules N. Therefore, doubling the number of molecules (at the same temperature) does not change the rms speed; it would, however, increase the pressure since P ∝ N for a fixed volume and temperature.

From the gas law PV = NkBT, temperature T = PV/(NkB) for a fixed number of molecules N. If both the volume V and pressure P are held constant, the right-hand side is fixed, so T is also fixed and cannot be changed. To raise the temperature of a gas, either its pressure or its volume (or both) must change.

At constant temperature the rms speed of the molecules does not change, since <K.E.> depends only on T. However, when the volume is reduced, the molecules have less distance to travel between successive collisions with the walls (the time between collisions, 2L/v, decreases as the box side L decreases), so they strike the walls more frequently. This higher rate of momentum transfer to the walls per unit area increases the pressure, consistent with P = (1/3)(N/V)m<v²> — pressure increases as V decreases.

A star is in a stable state (hydrostatic equilibrium) when the inward gravitational force is exactly balanced by the outward internal (thermal) pressure, as described by ΔP/Δr = −GM_rρ_r/r². If the gravitational force becomes greater than the internal pressure, the net inward force dominates and the star contracts, raising its density and temperature (possibly collapsing further into a neutron star or black hole). If the internal pressure becomes greater than the gravitational force, the net outward force dominates and the star expands, as happens when a star evolves into a red giant.

Since <K.E.> = (3/2)kBT, the average translational kinetic energy of gas molecules is directly proportional to the absolute temperature. As T approaches absolute zero (0 K), <K.E.> approaches zero, meaning the random thermal (translational) motion of the molecules effectively dies out.

The ink drop will gradually spread out and mix uniformly throughout the water due to Brownian motion. The water molecules are in continuous, rapid random motion and repeatedly collide with the ink particles, knocking them along a random zig-zag path. Over time, these random collisions cause the ink particles to diffuse outward from the region of high concentration (the drop) until they are spread evenly through the water.