"Thermodynamics" — Practice

Question 1

Which of internal energy, heat, work, pressure and temperature are state functions?

Accepted Answer

Internal energy, pressure and temperature are **state functions** — fixed by the current state alone.

Heat and work are **path functions** — they describe energy in transit and depend on the route taken between two states.

This is why a gas never "contains heat". It contains internal energy; heat is what crossed the boundary on the way.

Question 2

On a $P$–$V$ diagram, how do you tell an isothermal curve from an adiabatic one through the same point?

Accepted Answer

Differentiating $PV=c$ gives $\dfrac{dP}{dV}=-\dfrac{P}{V}$.

Differentiating $PV^{\gamma}=c$ gives $\dfrac{dP}{dV}=-\gamma\dfrac{P}{V}$.

So the **adiabatic is steeper, by exactly a factor of $\gamma$** at the shared point. Since $\gamma>1$ always, the adiabatic falls away faster.

Question 3

Why is $C_P$ greater than $C_V$ for any gas, and by how much?

Accepted Answer

At constant volume every joule supplied goes into internal energy. At constant pressure the gas also expands, so part of the heat is spent on work and more heat is needed for the same temperature rise.

The gap is exactly the expansion work per mole per degree, $R\Delta T$, so

$$C_P-C_V=R$$

The derivation needs $\Delta U=nC_V\Delta T$ to hold in the isobaric process — which is precisely why that theorem comes first.

Question 4

Two moles of a monatomic ideal gas are heated at constant pressure from $300\ 	ext{K}$ to $400\ 	ext{K}$. Find $Q$, $W$ and $\Delta U$. Take $R=8.31$.

Accepted Answer

$\Delta U=nC_V\Delta T=2\left(	frac32ight)(8.31)(100)=\mathbf{2493\ 	ext{J}}$

$W=P\Delta V=nR\Delta T=2(8.31)(100)=\mathbf{1662\ 	ext{J}}$

$Q=nC_P\Delta T=2\left(	frac52ight)(8.31)(100)=\mathbf{4155\ 	ext{J}}$

Check against the first law: $4155-1662=2493$. ✓

Note that only $60\%$ of the heat stayed in the gas, and $60\%$ is $C_V/C_P$ for a monatomic gas.

Question 5

A rigid diatomic gas at $300\ 	ext{K}$ is compressed adiabatically to one-eighth of its volume. Find the final temperature.

Accepted Answer

Use $TV^{\gamma-1}=	ext{constant}$ with $\gamma=1.4$ for a rigid diatomic gas:

$T_2=T_1\left(\dfrac{V_1}{V_2}ight)^{\gamma-1}=300	imes 8^{0.4}$

Since $8^{0.4}=2^{1.2}\approx 2.30$, the final temperature is about $\mathbf{690\ 	ext{K}}$.

More than double, with no heat supplied at all — which is exactly how a diesel engine ignites its fuel without a spark plug.

Question 6

A gas is taken clockwise around a rectangular cycle between pressures $P$ and $3P$ and volumes $V$ and $3V$. Find the net work and net heat per cycle.

Accepted Answer

The enclosed area is $(3P-P)(3V-V)=4PV$, and the cycle runs clockwise, so the net work is $\mathbf{+4PV}$.

Over a complete cycle the gas returns to its starting state, so $\Delta U=0$ and the first law gives $Q=W=\mathbf{4PV}$.

Positive work with heat absorbed means the device is a **heat engine**. Had the cycle run anticlockwise, both answers would flip sign and it would be a refrigerator.

Question 7

Find $\gamma$ for a mixture of $2$ moles of helium and $3$ moles of rigid diatomic oxygen.

Accepted Answer

$C_{V}=\dfrac{2\left(	frac32Right)+3\left(	frac52Right)}{5}=\dfrac{3R+7.5R}{5}=2.1R$

Mayer's relation gives $C_P=C_V+R=3.1R$, so

$\gamma=\dfrac{3.1}{2.1}\approx\mathbf{1.48}$

Averaging the individual gammas gives $\dfrac{2(1.67)+3(1.40)}{5}=1.51$ — wrong, because a ratio cannot be averaged, and close enough to sit in the options as a distractor.

Question 8

An ideal gas expands from $(P_0,V_0)$ to $\left(	frac{P_0}{2},2V_0ight)$, first isothermally and then along a straight line on the $P$–$V$ diagram. Compare the work done.

Accepted Answer

First note $P_0V_0=\left(	frac{P_0}{2}ight)(2V_0)$, so the two states are at the **same temperature** and $\Delta U=0$ on both paths.

Isothermal: $W=nRT\ln\dfrac{V_2}{V_1}=P_0V_0\ln 2=0.693\,P_0V_0$.

Straight line: the area under it is a trapezium of mean height $\dfrac{P_0+P_0/2}{2}=0.75P_0$ over a width $V_0$, giving $0.75\,P_0V_0$.

The straight path does **more** work. Same endpoints, same $\Delta U$, different $W$ — path dependence demonstrated in one comparison.

Question 9

An insulated container holds an ideal gas on one side of a partition and vacuum on the other. The partition is removed. Compare the final temperature with that of a reversible adiabatic expansion to the same final volume.

Accepted Answer

**Free expansion.** There is nothing to push against, so $W=0$; the container is insulated, so $Q=0$. The first law gives $\Delta U=0$, hence $\Delta T=0$ — **the temperature does not change**.

**Reversible adiabatic expansion.** The gas pushes a piston, so $W>0$ while $Q=0$, giving $\Delta U<0$ and the gas **cools**. For a doubling with $\gamma=1.4$, $\dfrac{T_2}{T_1}=\left(	frac12ight)^{0.4}\approx 0.76$.

Both are adiabatic; only the second is quasi-static. $PV^{\gamma}$ applies only to the second, and using it on the first is the most-set trap in the chapter.

Question 10

Why does a refrigerator not violate the Clausius statement of the second law, and how is that statement related to Kelvin–Planck?

Accepted Answer

Clausius forbids a process whose **sole result** is heat flowing from a colder body to a hotter one. A refrigerator does move heat that way, but not as its sole result — it also consumes external work, and that consumption is the second result the statement rules out.

Kelvin–Planck forbids a process whose sole result is the complete conversion of heat into work.

The two are logically equivalent: a device violating either could be combined with an ordinary engine or refrigerator to build a device violating the other. They stand or fall together.

Question 11

A rigid diatomic gas is taken along the path $PV^{1.2}=	ext{constant}$. Find its molar heat capacity and interpret the sign.

Accepted Answer

For a rigid diatomic gas $C_V=	frac52R$ and $\gamma=1.4$. With $n=1.2$:

$C=\dfrac52R+\dfrac{R}{1-1.2}=2.5R-5R=\mathbf{-2.5R}$

**Negative, and that is not an error.** The value $n=1.2$ lies strictly between $1$ (isothermal) and $\gamma=1.4$ (adiabatic), which is exactly the window where $C<0$.

*What it means physically:* during an expansion along this path the gas absorbs heat **and cools**. It does more work than the heat coming in, so the shortfall is taken out of its internal energy. Nothing is violated — $C$ is a property of the *process*, not a state function, so it is under no obligation to be positive.

*Check the endpoints of the window:* at $n=1$, $C	o\infty$ (heat in, no temperature change) and at $n=1.4$, $C=0$ (temperature change, no heat). A quantity passing from $+\infty$ to $0$ through that interval must go negative on the way.

Question 12

Two identical bodies, each of heat capacity $C=100\ 	ext{J/K}$, at $400\ 	ext{K}$ and $200\ 	ext{K}$, are placed in thermal contact and isolated from everything else. Find the final temperature and the total entropy change.

Accepted Answer

**Final temperature.** Heat lost equals heat gained, and the capacities are equal:

$T_f=\dfrac{T_1+T_2}{2}=\mathbf{300\ 	ext{K}}$

**Entropy.** Each body follows a reversible path at its own changing temperature, so $\Delta S=\displaystyle\int C\dfrac{dT}{T}=C\ln\dfrac{T_f}{T_i}$:

$\Delta S_{hot}=100\ln\dfrac{300}{400}=100(-0.288)=-28.8\ 	ext{J/K}$

$\Delta S_{cold}=100\ln\dfrac{300}{200}=100(0.405)=+40.5\ 	ext{J/K}$

$\Delta S_{total}=\mathbf{+11.8\ 	ext{J/K}}$

**Positive, as the second law requires** — the hot body's entropy fell, but the cold body's rose by more.

*The general proof is one line.* $\Delta S_{total}=C\ln\dfrac{T_f^{2}}{T_1T_2}$, and $T_f$ is the **arithmetic** mean while $\sqrt{T_1T_2}$ is the **geometric** mean. Since AM exceeds GM for any unequal pair, $T_f^{2}>T_1T_2$ and the logarithm is positive. Equality holds only when $T_1=T_2$ — that is, when nothing happens.

Practice — "Thermodynamics"

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