Degrees of Freedom and Law of Equipartition of Energy: notes and previous year questions

Answer: 7

The amount of gas is fixed, so $\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$, with the temperatures in kelvin: $27\ {}^{\circ}\mathrm{C} = 300\ \mathrm{K}$ and $77\ {}^{\circ}\mathrm{C} = 350\ \mathrm{K}$. The volumes can stay in $\mathrm{cm^{3}}$ and the pressures in atmospheres, since only ratios appear.

$p_{2} = p_{1} \times \dfrac{V_{1}}{V_{2}} \times \dfrac{T_{2}}{T_{1}} = 2 \times \dfrac{60}{20} \times \dfrac{350}{300} = 7\ \mathrm{atm}$

The trap is forgetting the heating: squeezing to a third of the volume alone gives $6\ \mathrm{atm}$, and the rise to $350\ \mathrm{K}$ adds the rest. Putting the Celsius values into the ratio, $\dfrac{77}{27}$, gives about $17\ \mathrm{atm}$ — far too much.

Answer: Option A

The two gases share one temperature, so in $v_{\mathrm{rms}} = \sqrt{\dfrac{3RT}{M}}$ only the molar mass differs, and $v_{\mathrm{rms}} \propto \dfrac{1}{\sqrt{M}}$:

$\dfrac{v_{\mathrm{rms}}^{\mathrm{Ar}}}{v_{\mathrm{rms}}^{\mathrm{Cl}}} = \sqrt{\dfrac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}} = \sqrt{\dfrac{70}{40}} = \sqrt{\dfrac{7}{4}} = \dfrac{\sqrt{7}}{2}$

The $2 : 1$ ratio of masses in the flask plays no part: how much of each gas there is does not change how fast its molecules move.

The trap is D, $\dfrac{2}{\sqrt{7}}$, the ratio upside down — it would make the heavier chlorine molecules the faster ones. B, $\dfrac{7}{4}$, forgets the square root, and C, $\dfrac{7}{2}$, lets the $2 : 1$ mass ratio in as well.

Answer: Option D

The air in the bubble is a fixed amount of gas, so $\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$. At the bottom it is pressed by the air above the pool and the water: $p_{1} = p_{0} + \rho g h = 10^{5} + 10^{3} \times 10 \times 5 = 1.5 \times 10^{5}\ \mathrm{Pa}$. At the surface, $p_{2} = 10^{5}\ \mathrm{Pa}$. In kelvin, $T_{1} = 290\ \mathrm{K}$ and $T_{2} = 300\ \mathrm{K}$.

$V_{2} = V_{1} \times \dfrac{p_{1}}{p_{2}} \times \dfrac{T_{2}}{T_{1}} = 2.9 \times 1.5 \times \dfrac{300}{290} = 4.5\ \mathrm{cm^{3}}$

The trap is C, $3.0\ \mathrm{cm^{3}}$, which allows for the warming but forgets that the pressure falls as the bubble rises. B, $4.2$, turns the temperature ratio upside down, and A, $2.0$, the pressure ratio. A rising bubble must grow: less pressure and more warmth both make it expand.

Answer: 3 atm

The volume is fixed, so $\dfrac{p}{T}$ is constant. In kelvin, $T_{1} = 300\ \mathrm{K}$ and $T_{2} = 177 + 273 = 450\ \mathrm{K}$, so

$p_{2} = 2.0 \times \dfrac{450}{300} = 3.0\ \mathrm{atm}$

The trap is using the Celsius values: $2.0 \times \dfrac{177}{27} \approx 13\ \mathrm{atm}$. A rise from $300$ to $450\ \mathrm{K}$ is only half as much again, and so is the pressure.

Answer: Option A

Helium is monatomic, $f = 3$, so $C_{V} = \dfrac{3}{2}R$ and $C_{P} = \dfrac{5}{2}R$. The heat is $nC_{P}\Delta T = 2 \times \dfrac{5}{2} \times 8.3 \times 10 = 415\ \mathrm{J}$: $249\ \mathrm{J}$ of it raises the internal energy and $166\ \mathrm{J}$ goes into pushing back the surroundings.

The trap is B, $249\ \mathrm{J}$, which uses $C_{V}$ — right at constant volume, but here the gas also expands. C, $166\ \mathrm{J}$, is only the extra part, $nR\Delta T$. D, $581\ \mathrm{J}$, uses $C_{P} = \dfrac{7}{2}R$, which belongs to a rigid diatomic gas.

Answer: 565.7 m/s

In kelvin the gas goes from $300\ \mathrm{K}$ to $600\ \mathrm{K}$, which doubles $T$. So $v_{\mathrm{rms}} = 400 \times \sqrt{2} \approx 566\ \mathrm{m/s}$.

The trap is the Celsius ratio: $400 \times \sqrt{\dfrac{327}{27}} \approx 1392\ \mathrm{m/s}$. Another is forgetting the square root and doubling the speed to $800\ \mathrm{m/s}$.