Thermodynamics

An insulated cylinder holds one mole of gas. In experiment A a piston is withdrawn infinitely slowly until the volume doubles. In experiment B a membrane bursts and the gas rushes into an equal evacuated chamber. Both are adiabatic. Do both cool the gas?

No. Experiment A cools it to $T/2^{\gamma -1}$. Experiment B leaves the temperature exactly unchanged.

The relation $P{V}^{\gamma }=$ constant is not the definition of an adiabatic process. It is derived by writing $dU=-PdV$ with $P$ being the gas's own pressure — which is only legitimate when the gas has a single well-defined pressure that matches the outside at every instant. That is the quasi-static, reversible case.

In a free expansion there is nothing to push against. $W=0$, and $Q=0$ because the walls are insulated, so

$\Delta U=0\Rightarrow \Delta T=0\text{(ideal gas)}$

Almost every hard problem in this chapter lives in that gap. Main gives you quasi-static processes and a formula per process. Advanced gives you a process that is not quasi-static, a process that is not one of the four, or a mixture of gases, and expects you to fall back on the first law itself.

1. Irreversible processes: use the first law, not the formula

When a gas expands against a constant external pressure inside insulated walls, the work is $P_{ext}\Delta V$, not $\int P_{gas}dV$. The first law gives directly

$n{C}_V\left(T_2-T_1\right)=-P_{ext}\left(V_2-V_1\right)$

which is one equation in $T_2$ once $V=nRT/P$ is substituted. No exponents appear anywhere.

Illustration 1

One mole of a monatomic ideal gas at $300$ K and $2$ atm expands adiabatically against a constant external pressure of $1$ atm until equilibrium. Find the final temperature, and compare with a reversible adiabatic expansion to the same final pressure.

With $P_2=P_{ext}=1$ atm and $C_V=\frac{3}{2}R$:

$\frac{3}{2}R\left(T_2-300\right)=-R\left(\frac{T_2}{1}-\frac{300}{2}\right)$

$1.5T_2-450=-T_2+150\Rightarrow T_2=240$ K

Reversibly, $T_2=300{\left(\frac{1}{2}\right)}^{2/5}=227.4$ K.

The irreversible route ends hotter. Less work is extracted, so less internal energy is spent — and the difference, $12.6$ K worth, is exactly the work that was wasted by pushing against a pressure lower than the gas's own.

Illustration 2

An insulated container is divided by a membrane. One mole of an ideal gas occupies volume $V$; the other half is evacuated. The membrane bursts. Find $\Delta U$, $W$, $\Delta T$ and $\Delta S$.

$W=0$ (nothing to push), $Q=0$ (insulated), so $\Delta U=0$ and $\Delta T=0$.

For entropy, invent a reversible isothermal path between the same endpoints:

$\Delta S=nR\ln \frac{V_2}{V_1}=R\ln 2=5.76$ J K${}^{-1}$

Entropy rose while nothing was exchanged with the surroundings, so the entropy of the universe increased by the same amount. That is what makes the process irreversible.

2. The molar heat capacity of an arbitrary process

$C_V$ and $C_P$ are two members of a continuous family. For any process, $C=\frac{1}{n}\frac{dQ}{dT}$, and the first law gives

$C=C_V+\frac{P}{n}\frac{dV}{dT}$

For the polytropic family $P{V}^k=$ constant this evaluates to

$C=C_V+\frac{R}{1-k}$

Setting $k=0,1,\gamma ,\infty$ recovers $C_P$, infinity, zero and $C_V$ in turn. The genuinely Advanced observation is what happens between $k=1$ and $k=\gamma$: there $C$ is negative.

For a monatomic gas at $k=1.5$, $C=\frac{3}{2}R-2R=-\frac{1}{2}R$. Supplying heat makes the temperature fall, because the gas does more work than the heat supplied. Nothing is violated — $C$ is a property of the path, not of the gas.

Illustration 3

A monatomic ideal gas is taken along a process for which $P=\alpha V$. Find its molar heat capacity, and the fraction of the supplied heat that becomes work.

$P=\alpha V$ means $P{V}^{-1}=$ constant, so $k=-1$:

$C=\frac{3}{2}R+\frac{R}{2}=2R$

$\Delta U=\frac{3}{2}R\Delta T$ of the $2R\Delta T$ supplied, so work is $\frac{1}{2}R\Delta T$ — one quarter of the heat becomes work.

Compare with an isobaric process, where $C=\frac{5}{2}R$ and the work fraction is $2/5$. Steepening the path towards the adiabatic always shifts the balance away from work and towards internal energy.

The bulk modulus of a gas, and why sound needs $\gamma$

Compressing a gas isothermally gives $B_{iso}=P$; compressing it adiabatically gives $B_{adia}=\gamma P$. Sound compressions are far too fast for heat to flow, so the adiabatic value is the correct one:

$v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma RT}{M}}$

Newton's isothermal calculation gives $280$ m s${}^{-1}$ in air; the measured value is $332$ m s${}^{-1}$, and $\sqrt{1.4}$ is exactly the missing factor.

3. Mixtures of gases

For a mixture, the internal energies add, so the mole-weighted heat capacities add:

$n{C}_{V,mix}=\sum n_i{C}_{V,i},C_{P,mix}=C_{V,mix}+R,{\gamma }_{mix}=\frac{C_{P,mix}}{C_{V,mix}}$

An equivalent form that is quicker in an exam is $\frac{n}{{\gamma }_{mix}-1}=\sum \frac{n_i}{{\gamma }_i-1}$.

Illustration 4

Find $\gamma$ for a mixture of $2$ mol of helium and $3$ mol of oxygen at room temperature.

$C_{V,mix}=\frac{2\left(\frac{3}{2}R\right)+3\left(\frac{5}{2}R\right)}{5}=\frac{3R+7.5R}{5}=2.1R$

$C_{P,mix}=3.1R\Rightarrow \gamma =\frac{3.1}{2.1}=1.476$

The answer must lie between $1.40$ and $1.67$, and closer to the diatomic end because oxygen supplies the majority of the moles. That bracket is the fastest check available on a mixture question.

4. Entropy: choose a reversible substitute

Entropy is a state function, so $\Delta S$ between two states is the same however the system actually got there. For an irreversible process, invent any reversible path joining the same endpoints and integrate $d{Q}_{rev}/T$ along it. For an ideal gas this gives

$\Delta S=n{C}_V\ln \frac{T_2}{T_1}+nR\ln \frac{V_2}{V_1}$

The surroundings, however, are not a state function shortcut — their entropy change is the actual heat they received divided by their own temperature.

Illustration 5

Equal masses of water at $300$ K and $400$ K are mixed in an insulated vessel. Find the entropy change of the universe, per unit $mc$.

$T_f=350$ K.

$\Delta S=mc\ln \frac{350}{300}+mc\ln \frac{350}{400}=mc\left(0.1542-0.1335\right)$

$\Delta S=0.0207mc$ J K${}^{-1}$, and positive.

It is always positive, because $T_f=\left(T_1+T_2\right)/2$ is the arithmetic mean while the product $T_1{T}_2$ under the logarithm is the geometric mean squared — and AM always exceeds GM unless the temperatures were already equal.

Illustration 6

A block at $600$ K is dropped into a large reservoir at $300$ K. Is $\Delta S_{universe}$ positive even though the block's entropy falls?

Block: $\Delta S=mc\ln \frac{300}{600}=-0.693mc$, negative.

Reservoir: it receives $mc(300)$ of heat at a fixed $300$ K, so $\Delta S=+1.000mc$.

Total $=+0.307mc$, positive.

The reservoir wins because it absorbs the heat at the lowest temperature involved. A negative entropy change for one part of a system is never by itself a contradiction.

5. Cycles, engines and refrigerators

For any cycle, $\Delta U=0$, so $W_{net}=Q_{net}$ and $W_{net}$ is the enclosed area, positive when traversed clockwise. Efficiency is defined only against the heat actually taken in:

$\eta =\frac{W_{net}}{Q_{in}}$

so the branches must be sorted into those with $Q>0$ and those with $Q<0$ before the ratio is formed. The Carnot bound and its two relatives are

${\eta }_C=1-\frac{T_C}{T_H},{\text{COP}}_{fridge}=\frac{T_C}{T_H-T_C},{\text{COP}}_{pump}=\frac{T_H}{T_H-T_C}$

and the last two differ by exactly one, since a heat pump delivers everything a refrigerator extracts plus the work paid for it.

Illustration 7

A monatomic gas is taken round a cycle: isochoric heating from $\left(P_0,V_0\right)$ to $\left(2P_0,V_0\right)$, isobaric expansion to $\left(2P_0,2V_0\right)$, isochoric cooling to $\left(P_0,2V_0\right)$, isobaric compression back. Find the efficiency.

$W_{net}=$ enclosed area $=P_0{V}_0$.

Heat in comes from the first two branches: $\frac{3}{2}{V}_0\left(2P_0-P_0\right)+\frac{5}{2}\left(2P_0\right)V_0=1.5P_0{V}_0+5P_0{V}_0=6.5P_0{V}_0$

$\eta =\frac{P_0{V}_0}{6.5P_0{V}_0}=15.4\%$

Compare the Carnot limit: the extremes are $T_{max}=4P_0{V}_0/R$ and $T_{min}=P_0{V}_0/R$, giving ${\eta }_C=75\%$. Rectangular cycles are spectacularly inefficient because most of their heat goes in at temperatures far below the maximum.

Illustration 8

An inventor claims an engine absorbing $1000$ J at $500$ K, rejecting $400$ J at $300$ K. Is the claim possible?

${\eta }_{claimed}=\frac{600}{1000}=60\%$, while ${\eta }_C=1-\frac{300}{500}=40\%$.

Impossible. Equivalently, $\Delta S_{universe}=-\frac{1000}{500}+\frac{400}{300}=-0.67$ J K${}^{-1}$, which is negative.

Use the entropy test, not the efficiency test, when the engine touches more than two reservoirs — it is the only version that generalises.

6. Thermal expansion and calorimetry, done carefully

Linear, areal and volume expansion are one relation seen three ways: $\beta =2\alpha$ and $\gamma =3\alpha$. The single fact that catches people is that a hole expands exactly as the removed material would have, so a cavity, a bore and a ring all grow, never shrink.

Three Advanced consequences follow.

Differential expansion. Bond two strips of different $\alpha$ and heat them; the pair must bend, because the longer strip has to sit on the outside. For strips each of thickness $t$,

$R\approx \frac{2t}{\left({\alpha }_2-{\alpha }_1\right)\Delta \theta }$

Apparent expansion of a liquid. A vessel expands too, so the level you observe reports only the difference:

${\gamma }_{app}={\gamma }_{liquid}-{\gamma }_{vessel}$

Density falls, and floating bodies ride higher or lower. Since ${\rho }^{\prime }=\rho /\left(1+\gamma \Delta \theta \right)$, warming a liquid more than the solid floating in it makes the solid sink deeper.

Illustration 9

A pendulum clock keeps correct time at $20^{\circ }$C. It is moved to a room at $35^{\circ }$C. Find the time it loses per day, with $\alpha =1.2\times 10^{-5}$ K${}^{-1}$.

$T\propto \sqrt{L}$, so the fractional change in period is half the fractional change in length:

$\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta =\frac{1}{2}\left(1.2\times 10^{-5}\right)(15)=9\times 10^{-5}$

Loss per day $=9\times 10^{-5}\times 86400=7.8$ s

It always loses, never gains, on warming, because a longer pendulum swings more slowly. The square root is what halves the coefficient, and forgetting it is the standard error here.

Illustration 10

$50$ g of ice at $0^{\circ }$C is added to $100$ g of water at $20^{\circ }$C. Find the final state. Take $L_f=80$ cal g${}^{-1}$ and $c=1$ cal g${}^{-1}$ K${}^{-1}$.

Heat available in cooling the water to $0^{\circ }$C: $100\times 1\times 20=2000$ cal.

Heat needed to melt all the ice: $50\times 80=4000$ cal.

Only $2000/80=25$ g of the ice melts. The final state is $125$ g of water and $25$ g of ice, together at $0^{\circ }$C.

Always test the phase change before assuming it completes. Writing a single mixing equation here would return a negative final temperature, which is the signal that the assumption was wrong.

7. Conduction as a resistance network

In the steady state, heat current through a slab is

$H=\frac{kA\Delta T}{L}=\frac{\Delta T}{R_{th}},R_{th}=\frac{L}{kA}$

which is Ohm's law with temperature as potential. Slabs end to end add their resistances; slabs side by side add their reciprocals. The junction temperature between two rods in series follows immediately by equating the currents.

Radial conduction replaces $L/A$ by an integral, because the area grows with radius. For a cylindrical shell $R_{th}=\frac{\ln \left(r_2/r_1\right)}{2\pi kL}$, and for a spherical shell $R_{th}=\frac{r_2-r_1}{4\pi k{r}_1{r}_2}$.

Illustration 11

Two rods of equal length and area, of conductivities $k$ and $3k$, are joined end to end. The free ends are held at $100^{\circ }$C and $0^{\circ }$C. Find the junction temperature.

$R_1=3R_2$, and the same current flows through both, so the temperature drops share in the ratio $3:1$.

The drop across the poorer conductor is $75^{\circ }$C, so the junction sits at

$T=100-75=25^{\circ }$C

The junction is always closer to the hot end of the better conductor. A quick sanity rule: most of the temperature falls where most of the resistance is.

Illustration 12

Ice of thickness $x$ has formed on a pond whose air temperature is $-\theta$. Show that the time to grow from $x_1$ to $x_2$ is proportional to $x_2^2-x_1^2$, and find how long doubling $5$ cm of ice takes relative to forming it.

Heat conducted through thickness $x$ freezes a layer $dx$:

$\frac{k\theta }{x}dt=\rho Ldx\Rightarrow t=\frac{\rho L}{2k\theta }\left(x_2^2-x_1^2\right)$

Forming the first $5$ cm goes as $25$; growing from $5$ to $10$ cm goes as $100-25=75$.

Doubling the thickness therefore takes three times as long as forming it.

8. Radiation: Stefan, Kirchhoff, Wien and Newton

A body at $T$ in surroundings at $T_s$ has a net radiated power

$P_{net}=e\sigma A\left(T^4-T_s^4\right)$

with $\sigma =5.67\times 10^{-8}$ W m${}^{-2}$ K${}^{-4}$. Kirchhoff's law states that absorptivity equals emissivity at each wavelength, so a good absorber is necessarily a good emitter — which is why a black surface both heats fastest in sunlight and cools fastest at night. Wien's law places the peak at

${\lambda }_{m}T=2.9\times 10^{-3}\text{m K}$

Newton's law of cooling is not an independent law. Put $T=T_s+\Delta T$ with $\Delta T\ll T_s$ and expand:

$T^4-T_s^4\approx 4T_s^3\Delta T$

so the loss becomes proportional to the excess temperature, and the cooling is exponential. That is exactly why Newton's law works for a cup of tea and fails for a filament.

Illustration 13

A body cools from $60^{\circ }$C to $50^{\circ }$C in $10$ minutes in surroundings at $20^{\circ }$C. How long to cool from $50^{\circ }$C to $40^{\circ }$C?

Using the average-excess form, $\frac{\Delta T}{t}=k\left(\overline{T}-T_s\right)$:

First stage: $\frac{10}{10}=k(55-20)=35k\Rightarrow k=\frac{1}{35}$

Second stage: $\frac{10}{t}=k(45-20)=\frac{25}{35}\Rightarrow t=14$ min

Each equal drop takes longer than the last, because the driving excess shrinks — the hallmark of exponential decay towards the surroundings.

Illustration 14

The Sun's spectrum peaks at $500$ nm. Estimate its surface temperature, and by what factor a star peaking at $250$ nm radiates more per unit area.

$T=\frac{2.9\times 10^{-3}}{500\times 10^{-9}}=5800$ K

Halving ${\lambda }_m$ doubles $T$, and Stefan's law then multiplies the emitted power per unit area by $2^4$:

The hotter star radiates 16 times as much per unit area.

Wien and Stefan work as a pair. Colour gives the temperature and temperature gives the intensity, which is how a star's luminosity is estimated without ever measuring its heat.

Summary

• $P{V}^{\gamma }$ describes only a reversible adiabatic. For anything irreversible, go back to $\Delta U=Q-W$.
• Free expansion: $W=0$, $Q=0$, so $\Delta T=0$ for an ideal gas — but $\Delta S=nR\ln \left(V_2/V_1\right)>0$.
• Adiabatic expansion against constant $P_{ext}$: $n{C}_V\left(T_2-T_1\right)=-P_{ext}\left(V_2-V_1\right)$, and the gas ends hotter than the reversible case.
• General molar heat capacity $C=C_V+\frac{R}{1-k}$ for $P{V}^k=$ constant, covering all four standard processes.
• For $1<k<\gamma$ the molar heat capacity is negative: heat goes in while the temperature falls.
• $B_{iso}=P$ but $B_{adia}=\gamma P$, so sound travels at $\sqrt{\gamma RT/M}$ — Laplace's correction to Newton.
• Mixtures: $n{C}_{V,mix}=\sum n_i{C}_{V,i}$, and ${\gamma }_{mix}$ must lie between the constituent values.
• Entropy is a state function: for an irreversible path, integrate $d{Q}_{rev}/T$ along any invented reversible path with the same endpoints.
• $\Delta S=n{C}_V\ln \left(T_2/T_1\right)+nR\ln \left(V_2/V_1\right)$; the surroundings' change uses the actual heat they received.
• Cycle: $W_{net}=$ enclosed area, and $\eta =W_{net}/Q_{in}$ counts only the branches with $Q>0$.
• ${\eta }_C=1-T_C/T_H$; ${\text{COP}}_{pump}={\text{COP}}_{fridge}+1$. Use the entropy test when more than two reservoirs appear.
• A hole expands like the material removed; ${\gamma }_{app}={\gamma }_{liquid}-{\gamma }_{vessel}$; a pendulum clock loses $\frac{1}{2}\alpha \Delta \theta$ per unit time.
• Conduction is Ohm's law: $R_{th}=L/kA$ in series and parallel, $\ln \left(r_2/r_1\right)/2\pi kL$ radially, and ice grows as $t\propto x^2$.
• Newton's cooling is Stefan's law linearised for small excess temperature, which is why it fails for hot bodies.