Entropy and Irreversibility: notes and previous year questions

Answer: Option D

The first law holds for every second of the process, so it can be written as rates: heat in per second $=$ rise of internal energy per second $+$ work done per second.

$100 = \dfrac{dU}{dt} + 75$, so $\dfrac{dU}{dt} = 25\ \mathrm{W}$.

The trap is A, $75\ \mathrm{W}$, the rate of doing work, and B, $100\ \mathrm{W}$, the whole heating rate — as if no work were done. C, $125\ \mathrm{W}$, adds the work to the heat, getting the sign of $W$ backwards: work done BY the system is energy leaving it.

Answer: Option D

At constant pressure, per mole and per kelvin: $\Delta Q = C_{P}\,\Delta T$, $\Delta U = C_{V}\,\Delta T$ and $\Delta W = R\,\Delta T$. For a diatomic gas $C_{V} = \dfrac{5}{2}R$ and $C_{P} = \dfrac{7}{2}R$, so

$\Delta Q : \Delta U : \Delta W = \dfrac{7}{2} : \dfrac{5}{2} : 1 = 7 : 5 : 2$

Check with the first law: $7 = 5 + 2$.

The trap is B, $5 : 3 : 2$, which is the monatomic gas ($\dfrac{5}{2} : \dfrac{3}{2} : 1$). A and C list the same numbers backwards, in the order $W : U : Q$ — and neither passes the check that $Q$ is the sum of the other two.

Answer: 350

At constant pressure $W = nR\,\Delta T$ and $Q = nC_{P}\,\Delta T$, so $\dfrac{Q}{W} = \dfrac{C_{P}}{R}$. With $\gamma = 1.4 = \dfrac{7}{5}$ the gas is diatomic, $C_{P} = \dfrac{7}{2}R$, and $\dfrac{Q}{W} = \dfrac{7}{2}$ (the same as $\dfrac{\gamma}{\gamma - 1} = \dfrac{1.4}{0.4} = 3.5$).

$Q = 3.5 \times 100 = 350\ \mathrm{J}$: $100\ \mathrm{J}$ goes into work and $250\ \mathrm{J}$ into internal energy.

The trap is $250\ \mathrm{J}$, which is $\Delta U$ and forgets the work itself. Another is $140\ \mathrm{J}$, $\gamma \times 100$, which treats $\gamma$ as the ratio of heat to work — that ratio is $\dfrac{C_{P}}{R}$, not $\dfrac{C_{P}}{C_{V}}$.

Answer: Option C

For a monatomic gas $C_{V} = \dfrac{3}{2}R$ and $C_{P} = \dfrac{5}{2}R$, so $\dfrac{W}{Q} = \dfrac{R}{C_{P}} = \dfrac{2}{5}$ and $W = \dfrac{2}{5} \times 500 = 200\ \mathrm{J}$. The other $300\ \mathrm{J}$ raises the internal energy.

The trap is A, $300\ \mathrm{J}$, which is $\Delta U$, not $W$. B, $500\ \mathrm{J}$, sends all the heat into work — but the gas warms up as it expands at constant pressure. D, $143\ \mathrm{J}$, uses the diatomic fraction $\dfrac{2}{7}$.

Answer: 415 J

$W = \dfrac{nR(T_{1} - T_{2})}{x - 1} = \dfrac{8.3 \times (300 - 400)}{-1 - 1} = \dfrac{-830}{-2} = 415\ \mathrm{J}$.

Check by area: $p = kV$ is a straight line through the origin, and the area under it is $\dfrac{k(V_{2}^{2} - V_{1}^{2})}{2} = \dfrac{p_{2}V_{2} - p_{1}V_{1}}{2} = \dfrac{R\,\Delta T}{2}$.

The trap is reading "$p$ proportional to $V$" as $x = +1$: written as $pV^{x}$ constant, $p = kV$ is $pV^{-1} = k$, so $x = -1$ (and $x = +1$ would divide by zero). Another is $nR\,\Delta T = 830\ \mathrm{J}$, the constant-pressure answer, twice too big.

Answer: Option C

Slope $-\gamma$ means $x = \gamma$: $pV^{\gamma}$ is constant, the adiabatic, with $C = 0$.

The trap is A, isothermal, which is $x = 1$ and slope $-1$. B, isobaric, is $x = 0$, a horizontal line. D, isochoric, keeps $\log V$ fixed — a vertical line, with no finite slope at all.