Chapter 11 · Thermodynamics

11.2 Specific Heat Capacities and Thermodynamic Processes

A gas needs more heat to warm it at constant pressure than at constant volume, $C_p-C_v=R$. Applying the first law to slow (quasi-static) changes gives the isothermal, adiabatic, isochoric, isobaric and cyclic processes — each a curve on the $p$–$V$ diagram whose area is the work.

A diesel engine has no spark plug. It squeezes air so quickly to about one-twentieth of its volume that the air heats up to several hundred degrees Celsius, and fuel sprayed into it ignites by itself. Air rising up the side of a mountain cools by about $10\ ^\circ\text{C}$ for every kilometre it climbs, with no refrigerator anywhere — and when it has cooled enough, its water vapour condenses into cloud. A gas cylinder warms slowly in the sun at fixed volume, while the air in a hot-air balloon is heated at constant pressure. Same first law, very different outcomes: what matters is how the gas is allowed to change.

In this section you will first see why a gas has two specific heat capacities, $C_v$ and $C_p$, and prove that for an ideal gas they differ by exactly $R$. Then you will describe the state of a gas by its state variables, introduce the idealised quasi-static process that can be drawn on a $p$–$V$ diagram, and work out the first-law account — $W$, $Q$ and $\Delta U$ — for the five processes that every engine, compressor and weather system is built from: isothermal, adiabatic, isochoric, isobaric and cyclic.

Specific heat capacities of a gas: $C_p-C_v=R$

In Chapter 10 the heat needed to warm a substance was $Q=mc\,\Delta T$, or, per mole, $Q=nC\,\Delta T$, with $C$ the molar specific heat capacity. For solids and liquids that is the whole story. A gas is different: how much heat it needs depends on whether it is allowed to expand while it is heated. If it expands, it does work on its surroundings, and that energy must also be supplied as heat. So for a gas we must say what is held fixed.

📘 Definition 11.5 — Molar specific heats $C_v$ and $C_p$

The molar specific heat at constant volume $C_v$ and the molar specific heat at constant pressure $C_p$ of a gas are the heats needed to raise the temperature of one mole by one kelvin with, respectively, the volume and the pressure held fixed:

$$C_v=\frac{1}{n}\left(\frac{\Delta Q}{\Delta T}\right)_{V},\qquad C_p=\frac{1}{n}\left(\frac{\Delta Q}{\Delta T}\right)_{p}\qquad(\text{unit } \text{J mol}^{-1}\text{K}^{-1}).$$

Their ratio is written $\gamma=\dfrac{C_p}{C_v}$ (the ratio of specific heats, or adiabatic index). Per kilogram, $c_v=C_v/M$ and $c_p=C_p/M$, where $M$ is the molar mass.

At constant volume the gas does no work ($W=0$), so all the heat goes into internal energy: $\Delta U=Q=nC_v\,\Delta T$. Now comes a key point. For an ideal gas $U$ depends only on $T$ (Definition 11.3), so the change in $U$ between two temperatures is the same whatever the process. Hence, for an ideal gas,

$$\Delta U=nC_v\,\Delta T\qquad\text{for every process, not only at constant volume.}$$

At constant pressure the gas expands and does work $W=p\,\Delta V$. From $pV=nRT$ at fixed $p$, $p\,\Delta V=nR\,\Delta T$. The first law then gives

$$nC_p\,\Delta T=Q=\Delta U+W=nC_v\,\Delta T+nR\,\Delta T .$$

📐 Result 11.4 — Mayer's relation

For an ideal gas, the molar specific heats at constant pressure and at constant volume differ by the gas constant:

$$C_p-C_v=R .$$

So $C_p\gt C_v$ and $\gamma\gt1$: at constant pressure the extra heat $nR\,\Delta T$ pays for the work done in expanding. Per kilogram, $c_p-c_v=\dfrac{R}{M}$.

The values of $C_v$ come from the way molecules store energy, which Chapter 12 — Kinetic Theory explains. For now, use the measured values for ideal gases near room temperature:

Ideal gas	$C_v$	$C_p$	$\gamma$	Examples
monatomic	$\tfrac32R$	$\tfrac52R$	$\tfrac53\approx1.67$	He, Ne, Ar
diatomic	$\tfrac52R$	$\tfrac72R$	$\tfrac75=1.40$	$\mathrm{N_2}$, $\mathrm{O_2}$, $\mathrm{H_2}$, air

For solids and liquids the expansion on heating is tiny, so the work $p\,\Delta V$ is negligible and $C_p\approx C_v$ — which is why Chapter 10 needed only one specific heat.

✏️ Example 1 — Heating helium two ways

$2.0\ \text{mol}$ of helium (monatomic) is warmed by $50\ \text{K}$. Find the heat needed (a) in a rigid container, (b) at constant pressure. (c) Where does the extra heat in (b) go?

💡 Show solution▶

(a) $Q_v=nC_v\,\Delta T=2.0\times\tfrac32(8.31)\times50=1.25\times10^3\ \text{J}$.

(b) $Q_p=nC_p\,\Delta T=2.0\times\tfrac52(8.31)\times50=2.08\times10^3\ \text{J}$.

(c) $\Delta U$ is the same in both ($1.25\ \text{kJ}$, same $\Delta T$). The extra $Q_p-Q_v=nR\,\Delta T=831\ \text{J}$ is the work the gas does pushing back its surroundings as it expands.

Check $\dfrac{Q_p}{Q_v}=\dfrac53=\gamma$ ✓.

✏️ Example 2 — Identifying a gas from its specific heats

It takes $291\ \text{J}$ to warm $0.50\ \text{mol}$ of a gas by $20\ \text{K}$ at constant pressure. (a) Find $C_p$, $C_v$ and $\gamma$. Is the gas monatomic or diatomic? (b) The gas is nitrogen ($M=28\ \text{g/mol}$). Find $c_p$ and $c_v$ per kilogram, and check that $c_p-c_v=R/M$.

💡 Show solution▶

(a) $C_p=\dfrac{291}{0.50\times20}=29.1\ \text{J mol}^{-1}\text{K}^{-1}$; $C_v=29.1-8.31=20.8\ \text{J mol}^{-1}\text{K}^{-1}$; $\gamma=\dfrac{29.1}{20.8}=1.40$ — diatomic ($C_p=\tfrac72R=29.1$).

(b) $c_p=\dfrac{29.1}{0.028}=1.04\times10^3\ \text{J kg}^{-1}\text{K}^{-1}$, $c_v=\dfrac{20.8}{0.028}=7.4\times10^2\ \text{J kg}^{-1}\text{K}^{-1}$.

Check $c_p-c_v=1039-742=297\ \text{J kg}^{-1}\text{K}^{-1}$ and $\dfrac{R}{M}=\dfrac{8.31}{0.028}=297$ ✓.

State variables, the equation of state and quasi-static processes

A gas in equilibrium in a cylinder is completely described, for thermodynamics, by a handful of numbers: its pressure, volume, temperature, amount and internal energy. These do not tell you the history of the gas — whether it was squeezed, heated or left alone — only its present condition.

📘 Definition 11.6 — State variables and the equation of state

The state variables of a system are the macroscopic quantities that describe an equilibrium state: for a gas, $p$, $V$, $T$, the amount $n$ (or mass) and the internal energy $U$. They have definite values only when the system is in equilibrium, and their values depend only on the state, not on how it was reached. Heat and work are not state variables.

The state variables are not independent. The relation between them is the equation of state; for an ideal gas it is $pV=nRT$. Fix any two of $p$, $V$, $T$ for a given amount of gas and the third — and hence $U$ — is determined, so an equilibrium state is one point on a $p$–$V$ diagram.

A variable that doubles when the system is doubled (two identical copies put side by side) is extensive — $V$, $n$, mass, $U$. One that stays the same is intensive — $p$, $T$, density.

A real change — a piston slammed in, a gas rushing into an empty container — passes through states that are not in equilibrium: the pressure is higher near the moving piston than far from it, and swirls and temperature differences appear. There is no single $p$ or $T$ for the gas, so the change cannot be drawn as a curve on the $p$–$V$ diagram. To use the diagram, we idealise.

📘 Definition 11.7 — Quasi-static process

A quasi-static process is one carried out so slowly that the system passes through a continuous sequence of equilibrium states: at every stage its pressure and temperature are uniform and differ only infinitesimally from those of the surroundings. It is drawn as a continuous curve on the $p$–$V$ diagram, and the work done is the area under that curve (Result 11.2).

In practice, move a piston in many tiny steps, or add weights to it a grain of sand at a time, and let the gas settle after each step; heat it by putting it in contact with a series of reservoirs, each only slightly hotter than the last. No real process is exactly quasi-static, but many come close. Four special quasi-static processes, and a fifth that returns to its start, are worth naming:

📘 Definition 11.8 — The basic thermodynamic processes

Isothermal — the temperature stays constant ($T=$ const). The system is in good thermal contact with a large reservoir and the change is slow.
Adiabatic — no heat is exchanged ($Q=0$). The system is insulated, or the change is so fast that heat has no time to flow.
Isochoric — the volume stays constant ($V=$ const, $W=0$): a gas heated in a rigid container.
Isobaric — the pressure stays constant ($p=$ const): a gas under a freely moving, weighted piston.
Cyclic — the system returns to its initial state, so every state variable, including $U$, comes back to its starting value.

Isothermal and adiabatic processes

Isothermal. For an ideal gas at constant $T$, $pV=nRT=$ constant (Boyle's law), so the path on the $p$–$V$ diagram is a rectangular hyperbola, an isotherm. Hotter isotherms lie further from the axes. Since $U$ depends only on $T$, $\Delta U=0$ and the first law gives $Q=W$: every joule of heat absorbed leaves again as work. The work is the area under the hyperbola $p=\dfrac{nRT}{V}$; adding up the strips $p\,dV$ uses the result $\displaystyle\int\frac{dV}{V}=\ln V$ (you will meet it in mathematics):

$$W=\int_{V_1}^{V_2}\frac{nRT}{V}\,dV=nRT\ln\frac{V_2}{V_1}.$$

📐 Result 11.5 — Isothermal process of an ideal gas

In a quasi-static isothermal change of an ideal gas at temperature $T$,

$$pV=\text{constant},\qquad \Delta U=0,\qquad Q=W=nRT\ln\frac{V_2}{V_1}=nRT\ln\frac{p_1}{p_2}.$$

In an expansion ($V_2\gt V_1$) the gas absorbs heat and does an equal amount of work; in a compression, work is done on it and it gives out an equal amount of heat.

✏️ Example 3 — A slow isothermal expansion

$1.0\ \text{mol}$ of an ideal gas at $300\ \text{K}$ is allowed to expand slowly, in contact with a large bath at $300\ \text{K}$, from $10\ \text{L}$ to $20\ \text{L}$. Find (a) the initial and final pressures, (b) the work done by the gas, (c) $\Delta U$ and the heat absorbed.

💡 Show solution▶

(a) $p_1=\dfrac{nRT}{V_1}=\dfrac{1.0\times8.31\times300}{0.010}=2.49\times10^5\ \text{Pa}$; doubling $V$ at constant $T$ halves $p$: $p_2=1.25\times10^5\ \text{Pa}$.

(b) $W=nRT\ln\dfrac{V_2}{V_1}=1.0\times8.31\times300\times\ln2=1.73\times10^3\ \text{J}$.

Check The area under the hyperbola must lie between the rectangles $p_2\Delta V=1.25\ \text{kJ}$ and $p_1\Delta V=2.49\ \text{kJ}$ ✓. Interactive 1 on "isothermal" with $V_2=20\ \text{L}$ integrates the area numerically and gets the same $1728\ \text{J}$.

Adiabatic. Now insulate the gas, $Q=0$. The first law gives $\Delta U=-W$: when the gas expands it does work at the expense of its internal energy and cools; when it is compressed, the work done on it raises its internal energy and it warms. That is the bicycle pump and the diesel engine. To find the path, apply the first law to a small step: $nC_v\,dT=-p\,dV$. With $p=\dfrac{nRT}{V}$ and $R=C_p-C_v$ this becomes $\dfrac{dT}{T}=-(\gamma-1)\dfrac{dV}{V}$, and adding up the steps (again with $\int\frac{dx}{x}=\ln x$) gives $\ln T+(\gamma-1)\ln V=$ constant, i.e. $TV^{\gamma-1}=$ constant. Eliminating $T$ with $pV=nRT$ gives the other forms.

📐 Result 11.6 — Adiabatic process of an ideal gas

In a quasi-static adiabatic change of an ideal gas ($Q=0$),

$$pV^\gamma=\text{constant},\qquad TV^{\gamma-1}=\text{constant},\qquad p^{1-\gamma}T^{\gamma}=\text{constant},$$

and the work done by the gas, from $\Delta U=-W$, is

$$W=nC_v(T_1-T_2)=\frac{p_1V_1-p_2V_2}{\gamma-1}=\frac{nR(T_1-T_2)}{\gamma-1}.$$

At any point an adiabat is steeper than the isotherm through it: $\left(\dfrac{dp}{dV}\right)_\text{adiabatic}=-\gamma\dfrac{p}{V}$, while $\left(\dfrac{dp}{dV}\right)_\text{isothermal}=-\dfrac{p}{V}$.

Why steeper? Expand a gas isothermally and its pressure falls only because the molecules are spread over a larger volume. Expand it adiabatically and, in addition, it cools — so its pressure falls faster. A sound wave squeezes and stretches air so quickly that the changes are adiabatic, which is why the speed of sound depends on $\gamma$ (Chapter 14 — Waves).

✏️ Example 4 — Ignition in a diesel engine

In a diesel engine, air ($\gamma=1.40$) at $300\ \text{K}$ and $1.0\times10^5\ \text{Pa}$ is compressed adiabatically to one-twentieth of its volume. Find (a) its final temperature, (b) its final pressure, (c) the work done on each mole of air.

💡 Show solution▶

(a) $T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1}=300\times20^{0.40}=300\times3.31=994\ \text{K}$, about $720\ ^\circ\text{C}$ — above the ignition temperature of diesel fuel.

(b) $p_2=p_1\left(\dfrac{V_1}{V_2}\right)^{\gamma}=1.0\times10^5\times20^{1.40}=6.6\times10^6\ \text{Pa}$, about $66$ atmospheres.

(c) $W=nC_v(T_1-T_2)=1\times\tfrac52(8.31)(300-994)=-1.44\times10^4\ \text{J}$: about $14\ \text{kJ}$ of work is done on each mole.

Check $p_2=p_1\dfrac{V_1}{V_2}\dfrac{T_2}{T_1}=1.0\times10^5\times20\times3.31=6.6\times10^6\ \text{Pa}$ ✓ (from $pV=nRT$). An isothermal compression would give only $20$ atmospheres.

✏️ Example 5 — Isothermal or adiabatic: which does more work?

$1.0\ \text{mol}$ of a monatomic ideal gas at $300\ \text{K}$ and $10\ \text{L}$ (the gas of Example 3) doubles its volume, this time adiabatically. Find (a) its final temperature and pressure, (b) the work it does. Compare with the isothermal expansion of Example 3.

💡 Show solution▶

(a) $\gamma=\tfrac53$: $T_2=300\times\left(\tfrac12\right)^{2/3}=300\times0.630=189\ \text{K}$; $p_2=2.49\times10^5\times\left(\tfrac12\right)^{5/3}=7.85\times10^4\ \text{Pa}$.

(b) $W=nC_v(T_1-T_2)=1.0\times\tfrac32(8.31)(300-189)=1.38\times10^3\ \text{J}$.

Compare Isothermal: $1.73\ \text{kJ}$; adiabatic: $1.38\ \text{kJ}$. The adiabat falls more steeply, so it encloses less area. In the isothermal case the bath keeps supplying heat; in the adiabatic case the work comes only from the gas's own internal energy.

Check $\dfrac{p_1V_1-p_2V_2}{\gamma-1}=\dfrac{2493-1571}{2/3}=1.38\times10^3\ \text{J}$ ✓.

$1.0\ \text{mol}$ of ideal gas starts at $300\ \text{K}$ and $10\ \text{L}$ (point 1). Choose a process and the final state; the state point moves along the path and the shaded area — computed by adding thin strips $p\,\Delta V$ — is the work done by the gas. Faint curves show the isotherm and the adiabat through point 1 for comparison. Check that $Q=\Delta U+W$ every time.

process:gas:final volume V₂:20.0 L

work by gas $W$—

change $\Delta U$—

heat in $Q$—

final state—

Isochoric, isobaric and cyclic processes

The two remaining straight-line processes are the easiest of all, because one of $W$ or $p$ is fixed.

📐 Result 11.7 — Isochoric and isobaric processes of an ideal gas

Isochoric ($V$ constant, a vertical line on the $p$–$V$ diagram): $\dfrac{p}{T}=$ constant and

$$W=0,\qquad Q=\Delta U=nC_v\,\Delta T .$$

Isobaric ($p$ constant, a horizontal line): $\dfrac{V}{T}=$ constant and

$$W=p\,\Delta V=nR\,\Delta T,\qquad \Delta U=nC_v\,\Delta T,\qquad Q=nC_p\,\Delta T .$$

In an isobaric expansion the fraction of the heat that becomes work is $\dfrac{W}{Q}=\dfrac{R}{C_p}=1-\dfrac1\gamma$: $\tfrac25$ for a monatomic gas, $\tfrac27$ for a diatomic one.

Cyclic processes. An engine must work over and over, so its working substance is taken round a closed loop on the $p$–$V$ diagram and back to its starting state. Over a complete cycle $\Delta U=0$, because $U$ is a state variable. On the expansion part of the loop the gas does positive work (area under the upper curve); on the compression part, negative work (area under the lower curve). What is left is the area inside the loop.

📐 Result 11.8 — Cyclic process

For a complete cycle,

$$\Delta U=0,\qquad Q_\text{net}=W_\text{net}=\text{area enclosed by the loop on the } p\text{–}V \text{ diagram}.$$

$W_\text{net}\gt0$ (the system does net work, like an engine) if the loop is traversed clockwise; $W_\text{net}\lt0$ (net work is done on the system) if it is traversed anticlockwise. Here $Q_\text{net}=Q_\text{in}-Q_\text{out}$, the heat absorbed on some parts of the cycle minus the heat given out on others.

✏️ Example 6 — A rectangular cycle

A monatomic ideal gas is taken round the cycle A → B → C → D → A, where A is $(1.0\times10^5\ \text{Pa},\ 2.0\ \text{L})$, B is $(3.0\times10^5\ \text{Pa},\ 2.0\ \text{L})$, C is $(3.0\times10^5\ \text{Pa},\ 5.0\ \text{L})$ and D is $(1.0\times10^5\ \text{Pa},\ 5.0\ \text{L})$. (a) Find the net work done by the gas per cycle. (b) Find $Q$ for each step. (c) How much heat is absorbed per cycle, how much is given out, and what fraction of the heat absorbed is turned into net work?

💡 Show solution▶

(a) The loop is a rectangle traversed clockwise (up, right, down, left): $W_\text{net}=\Delta p\,\Delta V=2.0\times10^5\times3.0\times10^{-3}=600\ \text{J}$.

(b) With $pV$ in joules ($p_AV_A=200$, $p_BV_B=600$, $p_CV_C=1500$, $p_DV_D=500$): A→B isochoric, $Q=\tfrac32\Delta(pV)=\tfrac32(400)=+600\ \text{J}$. B→C isobaric, $Q=\tfrac52p\,\Delta V=\tfrac52(900)=+2250\ \text{J}$. C→D isochoric, $Q=\tfrac32(-1000)=-1500\ \text{J}$. D→A isobaric, $Q=\tfrac52(-300)=-750\ \text{J}$.

(c) $Q_\text{in}=600+2250=2850\ \text{J}$; $Q_\text{out}=1500+750=2250\ \text{J}$; $\dfrac{W_\text{net}}{Q_\text{in}}=\dfrac{600}{2850}=0.21$ — only $21\%$.

Check $Q_\text{in}-Q_\text{out}=600\ \text{J}=W_\text{net}$ ✓. The four values of $\Delta U=\tfrac32\Delta(pV)$ are $+600$, $+1350$, $-1500$ and $-450\ \text{J}$, which add to zero ✓. Section 11.3 shows why no cycle can turn all of $Q_\text{in}$ into work.

An ideal gas goes round the rectangular cycle A → B → C → D → A, with A fixed at $(2.0\ \text{L},\ 1.0\times10^5\ \text{Pa})$. Set the top pressure and the largest volume. The state point runs round the loop; each step is coloured red while heat flows in and blue while heat flows out. On the right, the energy-flow diagram splits the heat absorbed per cycle into net work (green, the enclosed area) and heat rejected.

top pressure:3.0 × 10⁵ Palargest volume:5.0 Lgas:

heat absorbed $Q_\text{in}$—

heat rejected $Q_\text{out}$—

net work = area—

$W_\text{net}/Q_\text{in}$—

⚠️ Common mistakes

1. "Isothermal means no heat." No — isothermal means no change of temperature; an isothermal expansion absorbs heat equal to the work done. "No heat" is adiabatic.

2. "Adiabatic means constant temperature." The opposite: in an adiabatic change the temperature changes because no heat flows to keep it steady.

3. Using $\Delta U=nC_v\Delta T$ only at constant volume. For an ideal gas it holds for every process; only $Q=nC_v\Delta T$ is restricted to constant volume.

4. Celsius in $pV^\gamma$ or $TV^{\gamma-1}$. Temperatures must be in kelvin; pressures and volumes may be in any units as long as both sides use the same ones.

5. The sign of the net work in a cycle. Clockwise loop (with $V$ across and $p$ up): positive; anticlockwise: negative. Never add the areas under the curves — subtract.

6. $\ln$ versus $\log_{10}$. $W=nRT\ln\dfrac{V_2}{V_1}$ uses the natural logarithm; $\ln x=2.303\log_{10}x$.

Exercises

Take $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$; treat all gases as ideal, monatomic with $\gamma=\tfrac53$ and diatomic (and air) with $\gamma=1.40$, and all processes as quasi-static unless stated otherwise. $W$ is the work done by the gas, $\Delta U=Q-W$.

Specific heat capacities of gases.

1 Define $C_v$ and $C_p$ for a gas. Explain physically why $C_p\gt C_v$.
2 $3.0\ \text{mol}$ of argon is heated from $20\ ^\circ\text{C}$ to $70\ ^\circ\text{C}$. Find the heat needed (a) at constant volume, (b) at constant pressure.
3 $2.0\ \text{mol}$ of a diatomic gas absorbs $1.0\ \text{kJ}$ of heat at constant pressure. Find the rise in temperature, $\Delta U$ and $W$.
4 Show that when an ideal gas is heated at constant pressure, the fraction of the heat that is converted into work is $1-\dfrac1\gamma$. Evaluate it for monatomic and diatomic gases.
5 Find $c_p-c_v$ per kilogram for oxygen ($M=32\ \text{g/mol}$).
6 A gas has $c_p=14.3\ \text{kJ kg}^{-1}\text{K}^{-1}$ and $c_v=10.2\ \text{kJ kg}^{-1}\text{K}^{-1}$. Find $\gamma$ and the molar mass of the gas. Which gas is it likely to be?
7 Why is the difference between $C_p$ and $C_v$ negligible for solids and liquids?

💡 Show solutions▶

1 Heat per mole per kelvin with $V$ fixed ($C_v$) or $p$ fixed ($C_p$). At constant pressure the gas expands and does work $p\,\Delta V$, so extra heat is needed for the same rise in $U$ (same $\Delta T$).

2 $\Delta T=50\ \text{K}$. (a) $Q=3.0\times\tfrac32(8.31)(50)=1.87\times10^3\ \text{J}$; (b) $Q=3.0\times\tfrac52(8.31)(50)=3.12\times10^3\ \text{J}$.

3 $\Delta T=\dfrac{1000}{2.0\times\tfrac72(8.31)}=17\ \text{K}$ ($17.2$); $\Delta U=nC_v\Delta T=\tfrac57\times1000=714\ \text{J}$; $W=1000-714=286\ \text{J}$.

4 $\dfrac{W}{Q}=\dfrac{nR\,\Delta T}{nC_p\,\Delta T}=\dfrac{C_p-C_v}{C_p}=1-\dfrac{1}{\gamma}$: monatomic $1-\tfrac35=\tfrac25$; diatomic $1-\tfrac57=\tfrac27$.

5 $c_p-c_v=\dfrac{R}{M}=\dfrac{8.31}{0.032}=2.6\times10^2\ \text{J kg}^{-1}\text{K}^{-1}$.

6 $\gamma=\dfrac{14.3}{10.2}=1.40$ (diatomic). $\dfrac{R}{M}=4.1\times10^3$, so $M=\dfrac{8.31}{4100}=2.0\times10^{-3}\ \text{kg/mol}$ — hydrogen, $\mathrm{H_2}$.

7 They expand very little when heated, so the work $p\,\Delta V$ done against the atmosphere is tiny compared with $\Delta U$; almost all the heat goes into internal energy whichever quantity is held fixed.

State variables and quasi-static processes.

1 List the state variables of a gas. Which are extensive and which intensive?
2 A gas in equilibrium in a box is divided into two equal halves by a thin partition. Which of $p$, $V$, $T$, $n$, $U$ and density are halved?
3 What is a quasi-static process? Why is it an idealisation? Describe how you might carry out a nearly quasi-static compression.
4 A gas is allowed to rush into an evacuated vessel through a burst membrane. Can this change be drawn as a curve on the $p$–$V$ diagram? Explain.
5 Using the equation of state, find the amount of gas in a $2.0\ \text{L}$ flask at $1.5\times10^5\ \text{Pa}$ and $300\ \text{K}$.
6 Explain why a real isothermal process must be carried out slowly in a thin-walled metal container, while a real adiabatic process should be fast or well insulated.

💡 Show solutions▶

1 $p$, $V$, $T$, $n$ (or mass), $U$ (and density). Extensive: $V$, $n$, mass, $U$. Intensive: $p$, $T$, density.

2 $V$, $n$ and $U$ are halved (extensive); $p$, $T$ and density are unchanged (intensive).

3 A process so slow that the system passes through a continuous sequence of equilibrium states. Real changes always take finite time and create small non-uniformities, so it is an idealisation. Add tiny weights (grains of sand) one at a time to a piston, letting the gas settle after each.

4 No. During the rush the gas is far from equilibrium — pressure and temperature are not uniform — so it has no single $p$ or $T$. Only the initial and final equilibrium states can be plotted.

5 $n=\dfrac{pV}{RT}=\dfrac{1.5\times10^5\times2.0\times10^{-3}}{8.31\times300}=0.12\ \text{mol}$.

6 Isothermal: heat must flow in or out continuously to keep $T$ fixed, which needs good thermal contact (thin metal walls) and time. Adiabatic: no heat may flow, so either insulate the gas or make the change faster than heat can flow.

III

Isothermal and adiabatic processes.

1 $2.0\ \text{mol}$ of an ideal gas at $350\ \text{K}$ expands isothermally from $5.0\ \text{L}$ to $15.0\ \text{L}$. Find $W$, $\Delta U$ and $Q$.
2 $0.50\ \text{mol}$ of gas is compressed isothermally at $300\ \text{K}$ from $1.0\times10^5\ \text{Pa}$ to $4.0\times10^5\ \text{Pa}$. Find the work done on the gas and the heat it gives out.
3 A monatomic gas at $300\ \text{K}$ is compressed adiabatically to one-eighth of its volume. Find its final temperature and the factor by which its pressure increases.
4 Air at $5.0\times10^5\ \text{Pa}$ expands adiabatically from $2.0\ \text{L}$ to $6.0\ \text{L}$. Find the final pressure and the work done by the air.
5 Show that at any point of the $p$–$V$ diagram the slope of the adiabat is $\gamma$ times the slope of the isotherm.
6 A car tyre containing air at $27\ ^\circ\text{C}$ and $3.0\ \text{atm}$ (absolute) bursts. Treating the escaping air as expanding adiabatically to $1.0\ \text{atm}$, find its temperature.
7 The same gas expands from the same initial state to the same final volume, once isothermally and once adiabatically. In which case is more work done? Explain using the $p$–$V$ diagram.
8 $1.0\ \text{mol}$ of a diatomic gas at $400\ \text{K}$ expands adiabatically until its temperature is $300\ \text{K}$. Find the work done and the ratio $V_2/V_1$.

💡 Show solutions▶

1 $W=nRT\ln3=2.0\times8.31\times350\times1.099=6.39\times10^3\ \text{J}$; $\Delta U=0$; $Q=6.39\ \text{kJ}$ absorbed.

2 $W=nRT\ln\dfrac{p_1}{p_2}=0.50\times8.31\times300\times\ln\tfrac14=-1.73\times10^3\ \text{J}$: $1.73\ \text{kJ}$ of work is done on the gas, and since $\Delta U=0$ it gives out $1.73\ \text{kJ}$ of heat.

3 $T_2=300\times8^{2/3}=300\times4=1.2\times10^3\ \text{K}$; $\dfrac{p_2}{p_1}=8^{5/3}=32$.

4 $p_2=5.0\times10^5\times\left(\tfrac13\right)^{1.4}=1.07\times10^5\ \text{Pa}$. $W=\dfrac{p_1V_1-p_2V_2}{\gamma-1}=\dfrac{1000-644}{0.40}=8.9\times10^2\ \text{J}$.

5 Isotherm $pV=$ const: $V\,dp+p\,dV=0$, slope $-\dfrac pV$. Adiabat $pV^\gamma=$ const: $V^\gamma dp+\gamma pV^{\gamma-1}dV=0$, slope $-\gamma\dfrac pV$ — $\gamma$ times steeper.

6 $T_2=T_1\left(\dfrac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}=300\times\left(\tfrac13\right)^{0.286}=219\ \text{K}\approx-54\ ^\circ\text{C}$ — the blast of air feels icy.

7 Isothermal. From the same starting point the adiabat falls more steeply, so for the same $\Delta V$ it lies below the isotherm and the area under it is smaller (Examples 3 and 5: $1.73$ versus $1.38\ \text{kJ}$).

8 $W=nC_v(T_1-T_2)=\tfrac52(8.31)(100)=2.08\times10^3\ \text{J}$. $TV^{0.4}=$ const: $\dfrac{V_2}{V_1}=\left(\dfrac{400}{300}\right)^{2.5}=2.05$.

Isochoric, isobaric and cyclic processes; applications.

1 $0.40\ \text{mol}$ of helium in a rigid tank at $1.0\times10^5\ \text{Pa}$ and $300\ \text{K}$ is heated to $450\ \text{K}$. Find the final pressure and the heat supplied.
2 $1.0\ \text{mol}$ of a diatomic gas at a constant $2.0\times10^5\ \text{Pa}$ expands from $10\ \text{L}$ to $15\ \text{L}$. Find $W$, $\Delta U$, $Q$ and the rise in temperature.
3 A gas is taken round the triangle A $(1.0\ \text{L},\ 1.0\times10^5\ \text{Pa})$ → C $(1.0\ \text{L},\ 3.0\times10^5\ \text{Pa})$ → B $(3.0\ \text{L},\ 1.0\times10^5\ \text{Pa})$ → A along straight lines. Find the net work done by the gas and the net heat absorbed per cycle.
4 If the cycle of part 3 is traversed in the opposite direction, what are the net work and net heat? What does the sign mean?
5 (Modelling) Dry air rising up a mountain expands adiabatically from $1.00\times10^5\ \text{Pa}$ and $300\ \text{K}$ at the foot to $0.80\times10^5\ \text{Pa}$ at the summit. By how much does it cool?
6 (Modelling) In a quick stroke, a bicycle pump compresses air adiabatically from $1.0\ \text{atm}$ and $300\ \text{K}$ to $3.0\ \text{atm}$. Find the temperature of the compressed air.
7 A diatomic gas goes round the rectangle A $(1.0\times10^5\ \text{Pa},\ 1.0\ \text{L})$ → B $(2.0\times10^5\ \text{Pa},\ 1.0\ \text{L})$ → C $(2.0\times10^5\ \text{Pa},\ 3.0\ \text{L})$ → D $(1.0\times10^5\ \text{Pa},\ 3.0\ \text{L})$ → A. Find $W_\text{net}$, $Q_\text{in}$, $Q_\text{out}$ and $W_\text{net}/Q_\text{in}$.

💡 Show solutions▶

1 $p\propto T$: $p_2=1.5\times10^5\ \text{Pa}$. $Q=nC_v\Delta T=0.40\times\tfrac32(8.31)(150)=7.5\times10^2\ \text{J}$.

2 $W=2.0\times10^5\times5.0\times10^{-3}=1.0\times10^3\ \text{J}$; $\Delta T=\dfrac{W}{nR}=120\ \text{K}$; $\Delta U=\tfrac52W=2.5\times10^3\ \text{J}$; $Q=\tfrac72W=3.5\times10^3\ \text{J}$.

3 A → C is up, C → B down to the right, B → A left: clockwise. $W_\text{net}=\tfrac12(2.0\times10^{-3})(2.0\times10^5)=200\ \text{J}$; $\Delta U=0$, so $Q_\text{net}=200\ \text{J}$ absorbed.

4 $W_\text{net}=-200\ \text{J}$ and $Q_\text{net}=-200\ \text{J}$: net work is done on the gas each cycle and it gives out that much heat — the way a refrigerator or heat pump runs (Section 11.3).

5 $T_2=300\times(0.80)^{0.286}=281\ \text{K}$: it cools by about $19\ \text{K}$ — the reason mountain tops are cold and clouds form over hills.

6 $T_2=300\times3^{0.286}=411\ \text{K}$ (about $137\ ^\circ\text{C}$) — in practice less, as heat leaks to the barrel, but the pump does get hot.

7 $pV$ in joules: $100$, $200$, $600$, $300$. $W_\text{net}=1.0\times10^5\times2.0\times10^{-3}=200\ \text{J}$. A→B $Q=\tfrac52(100)=250\ \text{J}$; B→C $Q=\tfrac72(400)=1400\ \text{J}$; C→D $Q=\tfrac52(-300)=-750\ \text{J}$; D→A $Q=\tfrac72(-200)=-700\ \text{J}$. $Q_\text{in}=1650\ \text{J}$, $Q_\text{out}=1450\ \text{J}$, $\dfrac{W_\text{net}}{Q_\text{in}}=0.12$.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Define the molar specific heats $C_v$ and $C_p$ of a gas and show that for an ideal gas $C_p-C_v=R$.

💡 Show solution▶

Definitions $C_v$ ($C_p$): heat needed to raise the temperature of one mole by $1\ \text{K}$ at constant volume (pressure). [1]

Constant V $W=0$, so $Q=\Delta U=nC_v\Delta T$. For an ideal gas $U$ depends only on $T$, so $\Delta U=nC_v\Delta T$ for any process. [1]

Constant p $Q=nC_p\Delta T=\Delta U+p\Delta V=nC_v\Delta T+nR\Delta T$ (from $pV=nRT$), so $C_p-C_v=R$. [1]

Exam tip The step "$\Delta U$ is the same in both processes because $U$ depends only on $T$" is what earns the middle mark — write it explicitly.

E2. NEET When a diatomic ideal gas is heated at constant pressure, the fraction of the heat supplied that increases its internal energy is (a) $\tfrac27$ (b) $\tfrac57$ (c) $\tfrac35$ (d) $\tfrac25$

💡 Show solution▶

Idea $\dfrac{\Delta U}{Q}=\dfrac{nC_v\Delta T}{nC_p\Delta T}=\dfrac{1}{\gamma}=\dfrac{5/2}{7/2}$.

Answer (b) $\tfrac57$.

Exam tip (a) is the fraction that becomes work; (c) is the monatomic answer. Remember the pair: monatomic $\tfrac35$ and $\tfrac25$, diatomic $\tfrac57$ and $\tfrac27$.

E3. JEE Main (Numerical value) $2.0\ \text{mol}$ of a monatomic ideal gas is compressed adiabatically, and its temperature rises from $300\ \text{K}$ to $500\ \text{K}$. Find the work done on the gas, in joules, to the nearest integer ($R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$).

💡 Show solution▶

Idea $Q=0$, so work done on the gas $=-W=\Delta U=nC_v\Delta T$.

Numbers $2.0\times\tfrac32\times8.31\times200=4986\ \text{J}$.

Answer $4986$.

Exam tip No volumes or pressures are needed: in an adiabatic change of an ideal gas, the work is fixed by the temperature change alone.

E4. JEE Advanced (One or more correct) A monatomic ideal gas at $(p_0,V_0)$ is expanded to volume $2V_0$, separately, by an isothermal, an adiabatic and an isobaric process. Which statements are correct? (A) $W_\text{isobaric}\gt W_\text{isothermal}\gt W_\text{adiabatic}$. (B) The final temperature is lowest after the adiabatic expansion. (C) $W_\text{adiabatic}=\tfrac32p_0V_0\left(1-2^{-2/3}\right)$. (D) In the isobaric expansion the internal energy increases by $p_0V_0$.

💡 Show solution▶

(A) $W_\text{isobaric}=p_0V_0$, $W_\text{isothermal}=p_0V_0\ln2=0.693\,p_0V_0$, $W_\text{adiabatic}=0.555\,p_0V_0$ (from (C)) ✓ — the paths lie one below the other.

(B) Isobaric: $T$ doubles; isothermal: $T_0$; adiabatic: $T_0\,2^{-2/3}=0.63\,T_0$ ✓.

(C) $p_2V_2=p_0\,2^{-5/3}\cdot2V_0=2^{-2/3}p_0V_0$; $W=\dfrac{p_0V_0-p_2V_2}{\gamma-1}=\tfrac32p_0V_0\left(1-2^{-2/3}\right)$ ✓.

(D) $\Delta U=\tfrac32\Delta(pV)=\tfrac32p_0V_0$ ✗ ($p_0V_0$ is the work).

Answer (A), (B), (C).

Exam tip Sketch the three curves from one point: horizontal (isobaric) on top, then the isotherm, then the steeper adiabat. Areas and final temperatures follow from the picture.

E5. NEET JEE Main A gas is taken round a rectangle on the $p$–$V$ diagram with corners at pressures $1.0\times10^5$ and $3.0\times10^5\ \text{Pa}$ and volumes $2.0$ and $4.0\ \text{L}$, in the anticlockwise direction. The net work done by the gas per cycle is (a) $+400\ \text{J}$ (b) $-400\ \text{J}$ (c) $+1200\ \text{J}$ (d) zero, because $\Delta U=0$

💡 Show solution▶

Idea $|W_\text{net}|=$ area $=2.0\times10^5\times2.0\times10^{-3}=400\ \text{J}$; anticlockwise, so negative.

Answer (b) $-400\ \text{J}$ — net work is done on the gas, and it gives out $400\ \text{J}$ of heat per cycle.

Exam tip (d) confuses $\Delta U=0$ with $W=0$; over a cycle $Q_\text{net}=W_\text{net}$, both non-zero. (c) is the area under the top side alone.

E6. Board · Case study In a diesel engine air is drawn in at $27\ ^\circ\text{C}$ and $1.0\times10^5\ \text{Pa}$ and compressed rapidly to one-sixteenth of its volume ($\gamma=1.4$). (i) Why can the compression be treated as adiabatic? [1] (ii) Find the temperature of the compressed air. [2] (iii) Find its pressure. [1]

💡 Show solution▶

(i) The compression stroke lasts a few hundredths of a second — too short for appreciable heat to flow through the cylinder walls, so $Q\approx0$. [1]

(ii) $TV^{\gamma-1}=$ const: $T_2=300\times16^{0.4}=300\times3.03=909\ \text{K}$ (about $636\ ^\circ\text{C}$). [2]

(iii) $p_2=1.0\times10^5\times16^{1.4}=4.9\times10^6\ \text{Pa}$ (about $48$ atm). [1]

Exam tip Convert $27\ ^\circ\text{C}$ to $300\ \text{K}$ before using $TV^{\gamma-1}=$ const; check with $p_2=p_1\dfrac{V_1}{V_2}\dfrac{T_2}{T_1}=16\times3.03\times10^5$ ✓.

🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How fast does rising air cool — and can an adiabatic expansion predict the height of the cloud base on a summer afternoon?

Why it matters. Meteorologists, glider and paraglider pilots, and air-pollution scientists all rely on the cooling of rising air. A parcel of air carried upwards meets lower pressure, expands almost adiabatically (air is a poor conductor and the rise takes minutes) and cools. When it has cooled to its dew point, cloud forms. Whether the surrounding air cools faster or slower with height decides whether the atmosphere is stable or produces thunderstorms.

What to do.

Apply the first law to one kilogram of rising air. Show that, with $dp=-\rho g\,dz$ for the atmosphere and an adiabatic expansion, the temperature falls with height at the rate $\dfrac{dT}{dz}=-\dfrac{g}{c_p}$ (the dry adiabatic lapse rate). Evaluate it with $c_p=1005\ \text{J kg}^{-1}\text{K}^{-1}$.
Check your result with $p^{1-\gamma}T^\gamma=$ const: take $1.00\times10^5\ \text{Pa}$ at the ground and about $0.89\times10^5\ \text{Pa}$ at $1\ \text{km}$.
Download a recent weather-balloon (radiosonde) record for a station near you from a public meteorological archive. Plot the measured temperature against height for the lowest $3\ \text{km}$ and compare its slope with your lapse rate.
A rising parcel's dew point falls by only about $2\ \text{K}$ per km. Using the ground temperature and dew point from the record, predict the height at which cloud should form, and compare with the reported cloud base.

Tools: Spreadsheet, public radiosonde data, a weather report with temperature and dew point.

Hand in: Your derivation, the lapse-rate graph with the measured profile and the dry adiabat, your cloud-base prediction and a paragraph explaining any difference.

Going further. Above the cloud base the parcel cools more slowly, because condensing vapour releases latent heat. Find out about the "moist adiabatic lapse rate" and why it allows thunderclouds to tower to $10\ \text{km}$ or more.

Instructor note: $\dfrac{g}{c_p}=\dfrac{9.8}{1005}=9.75\times10^{-3}\ \text{K/m}$, about $9.8\ \text{K}$ per km. With $\gamma=1.4$, $T_2=T_1(p_2/p_1)^{0.286}$ gives, from $300\ \text{K}$ and $0.89$ of the pressure, a fall of $9.8\ \text{K}$ — consistent. Measured average profiles fall only about $6.5\ \text{K/km}$ (the environment is not a rising parcel, and moisture matters). The cloud-base estimate is roughly $125\ \text{m}$ per kelvin of the difference between temperature and dew point at the ground (from $9.8-2\approx8\ \text{K/km}$).

🧪 Self-check quiz

0/5 correct

Q1. In an isothermal expansion of an ideal gas,

Q2. When an ideal gas expands adiabatically, its temperature

Q3. For a monatomic ideal gas, $C_p$ is

Q4. The net work done by a gas in one complete cycle equals

Q5. $1.0\ \text{mol}$ of a monatomic ideal gas in a rigid container is warmed by $10\ \text{K}$. The heat absorbed is about

← 11.1 Thermal Equilibrium, the Zeroth Law and the First Law 11.3 The Second Law, Reversible Processes and the Carnot Engine →