"Thermodynamics" — Practice

Question 1

Two moles of a diatomic ideal gas at $400$ K and $4$ atm expand adiabatically against a constant external pressure of $1$ atm until equilibrium. Find the final temperature and the work done.

Accepted Answer

With $C_V=	frac52R$ and $P_2=P_{ext}=1$ atm:

$2\left(	frac52Right)\left(T_2-400ight)=-2R\left(\dfrac{T_2}{1}-\dfrac{400}{4}ight)$

$5T_2-2000=-2T_2+200\qquad\Rightarrow\qquad T_2=\dfrac{2200}{7}=314.3$ K

$W=-\Delta U=-2\left(	frac52Right)\left(314.3-400ight)=5R(85.7)=3562$ J

Question 2

An insulated cylinder contains $1$ mol of an ideal gas at $300$ K in half its volume, the other half evacuated. The partition is removed. Find $\Delta T$, $\Delta S_{gas}$ and $\Delta S_{universe}$.

Accepted Answer

$W=0$ and $Q=0$, so $\Delta U=0$ and $\Delta T=0$ — the gas stays at $300$ K.

$\Delta S_{gas}=nR\ln\dfrac{V_2}{V_1}=R\ln2=5.76$ J K$^{-1}$

The surroundings exchanged nothing, so $\Delta S_{surr}=0$ and

$\Delta S_{universe}=+5.76$ J K$^{-1}$

which is why the process cannot run backwards.

Question 3

A gas expands adiabatically and reversibly from $\left(P_0,V_0\right)$ to $2V_0$, and separately expands adiabatically and irreversibly to the same final volume against a constant pressure. Which ends at the higher temperature, and why?

Accepted Answer

The **irreversible** expansion ends hotter.

In the reversible case the gas pushes against its own pressure at every instant, which is the largest pressure it can push against. It therefore does the maximum possible work.

Since both are adiabatic, $\Delta U=-W$, so the larger work means the larger drop in internal energy and the lower final temperature.

The reversible path is the extreme case, and every irreversible route between the same volumes extracts less work.

Question 4

A monatomic ideal gas follows $T=\alpha V^{2}$. Find its molar heat capacity, and the fraction of the heat supplied that appears as work.

Accepted Answer

$PV=nRT=nR\alpha V^{2}$, so $P\propto V$, i.e. $PV^{-1}=$ constant and $k=-1$.

$C=C_V+\dfrac{R}{1-(-1)}=	frac32R+	frac12R=2R$

Of $2R\,\Delta T$ supplied, $	frac32R\,\Delta T$ goes into internal energy, so work is $	frac12R\,\Delta T$.

The work fraction is $\dfrac{1}{4}$.

Question 5

For what range of the polytropic index $k$ does a monatomic ideal gas have a negative molar heat capacity? Interpret the result physically.

Accepted Answer

$C= frac32R+\dfrac{R}{1-k}<0$ For $k>1$ write $C= frac32R-\dfrac{R}{k-1}$, which is negative when $\dfrac{R}{k-1}> frac32R$, i.e. $k-1< frac23$. So $1

Question 6

Find $C_V$, $C_P$ and $\gamma$ for a mixture of $3$ mol of argon and $2$ mol of nitrogen at room temperature, and state the bracket the answer must fall in.

Accepted Answer

$C_{V,mix}=\dfrac{3\left(	frac32Right)+2\left(	frac52Right)}{5}=\dfrac{4.5R+5R}{5}=1.9R$

$C_{P,mix}=2.9R$

$\gamma=\dfrac{2.9}{1.9}=1.526$

It must lie between $1.40$ and $1.67$, and nearer the monatomic end because argon supplies more moles — which it does.

Question 7

One mole of an ideal gas is heated at constant volume from $300$ K to $600$ K, then expanded isothermally to twice its volume. Find the total entropy change. Take $C_V=	frac32R$.

Accepted Answer

Stage 1: $\Delta S_1=	frac32R\ln2=0.693\left(1.5Right)=1.04R$

Stage 2: $\Delta S_2=R\ln2=0.693R$

$\Delta S=1.73R=14.4$ J K$^{-1}$

Applying the combined formula between the initial and final states gives the same value, since $S$ is a state function.

Question 8

An engine is claimed to absorb $800$ J at $600$ K, reject $300$ J at $400$ K and reject a further $100$ J at $300$ K. Is it possible?

Accepted Answer

$W=800-300-100=400$ J, so $\eta=50\%$.

$\Delta S_{universe}=-\dfrac{800}{600}+\dfrac{300}{400}+\dfrac{100}{300}$

$=-1.333+0.750+0.333=-0.25$ J K$^{-1}$

Negative, so the engine is **impossible**. The entropy test is the version that generalises beyond two reservoirs.

Question 9

A Carnot refrigerator maintains $-13^{\circ}$C in a room at $27^{\circ}$C. Find its coefficient of performance and the work needed to remove $1000$ J of heat.

Accepted Answer

$T_C=260$ K, $T_H=300$ K.

$	ext{COP}=\dfrac{260}{40}=6.5$

$W=\dfrac{Q_C}{	ext{COP}}=\dfrac{1000}{6.5}=154$ J

Used as a heat pump between the same reservoirs, the coefficient would be $7.5$, larger by exactly one.

Question 10

Three rods of equal length and area, of conductivities $k$, $2k$ and $3k$, are joined end to end. The free ends are at $100^{\circ}$C and $0^{\circ}$C. Find both junction temperatures.

Accepted Answer

Resistances are in the ratio $1:	frac12:	frac13$, that is $6:3:2$ after scaling.

The total is $11$ parts, so the drops are $\dfrac{6}{11}(100)=54.5$, $\dfrac{3}{11}(100)=27.3$ and $\dfrac{2}{11}(100)=18.2$ degrees.

First junction: $100-54.5=45.5^{\circ}$C

Second junction: $45.5-27.3=18.2^{\circ}$C

Question 11

A lake is covered with $4$ cm of ice at an air temperature of $-10^{\circ}$C. Find the ratio of the time already taken to the further time needed to reach $12$ cm.

Accepted Answer

$t\propto x^{2}$, so forming $4$ cm goes as $16$.

Growing from $4$ cm to $12$ cm goes as $144-16=128$.

Ratio $=\dfrac{16}{128}=\dfrac{1}{8}$

The further growth takes eight times as long, because each new layer must conduct its latent heat through everything already formed.

Question 12

A body cools from $80^{\circ}$C to $64^{\circ}$C in $5$ minutes in surroundings at $24^{\circ}$C. Find the time to cool from $64^{\circ}$C to $52^{\circ}$C, and state when this method stops being valid.

Accepted Answer

First stage: $\dfrac{16}{5}=k\left(72-24ight)=48k\qquad\Rightarrow\qquad k=\dfrac{1}{15}$

Second stage: $\dfrac{12}{t}=k\left(58-24ight)=\dfrac{34}{15}\qquad\Rightarrow\qquad t=5.3$ min

The method assumes $T^{4}-T_s^{4}\approx4T_s^{3}\Delta T$, so it fails once the excess temperature is no longer small compared with the absolute temperature of the surroundings.

Practice — "Thermodynamics"

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