Why a Gas Needs More Heat to Warm Up When It Is Allowed to Expand

A pressure cooker whistles, a car engine turns fuel into motion, and an air conditioner pumps heat out of a room. Thermodynamics connects heat, work and internal energy through a few laws that apply to every such machine.

This lesson covers the zeroth and first laws with isothermal and adiabatic processes, the two specific heats of a gas and Mayer's relation, and the second law with heat engines and refrigerators.

**The zeroth law says two bodies each in thermal equilibrium with a third are in equilibrium with each other, defining temperature; the first law says $\Delta Q=\Delta U+\Delta W$, so in an isothermal process heat becomes work, and in an adiabatic process work comes entirely from internal energy.

Zeroth law. If A and B are each in thermal equilibrium with C, they are in thermal equilibrium with each other — the basis of every thermometer.

First law:**

$$\Delta Q=\Delta U+\Delta W$$

- $\Delta W$ is the work done by the system, $p\Delta V$ at constant pressure — unlike chemistry's convention, which counts work done on the system

Isothermal process — constant temperature, pV constant:

- For an ideal gas $\Delta U=0$, so $\Delta Q=\Delta W$
- $W=2.303nRT\log \frac{V_2}{V_1}$

Adiabatic process — no heat exchanged, $p{V}^{\gamma }$ constant:

- $\Delta Q=0$, so $\Delta W=-\Delta U$: expansion cools a gas and compression heats it
- $W=\frac{nR(T_1-T_2)}{\gamma -1}$, and $T{V}^{\gamma -1}$ is constant

Worked example 1 — isothermal expansion. 1.0 mol of gas at 400 K expands isothermally from 10 L to 20 L:

$$W=2.303\times 1.0\times 8.31\times 400\times \log 2=2.30\times 10^3\text{ J}$$

Worked example 2 — adiabatic compression. Air, with $\gamma =1.4$, at 300 K is compressed adiabatically to one-eighth of its volume:

$$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300\times 8^{0.4}=689\text{ K}$$

An everyday example. Air rushing out of a burst tyre at a roadside puncture shop feels cold, because the rapid, nearly adiabatic expansion uses up the gas's internal energy.

The substance. An adiabatic curve is steeper than an isothermal one on a p-V graph — pressure changes faster because temperature changes too.

**$C_v$ is the heat needed to raise one mole of gas by 1 K at constant volume, $C_p$ is the heat needed at constant pressure, where the gas expands, and Mayer's relation states $C_p-C_v=R$.

Why $C_p$ is larger:**

- At constant volume no work is done, so all the heat raises internal energy: $n{C}_v\Delta T=\Delta U$
- At constant pressure the gas also does work $p\Delta V$ while expanding, so extra heat is needed

Deriving Mayer's relation. At constant pressure, $n{C}_p\Delta T=n{C}_v\Delta T+p\Delta V$. For an ideal gas $p\Delta V=nR\Delta T$, so

$$C_p-C_v=R$$

Ratio of specific heats. $\gamma =\frac{C_p}{C_v}$ is about 1.67 for monatomic, 1.40 for diatomic and 1.33 for polyatomic gases.

Worked example. For a diatomic gas, $C_v=\frac{5}{2}R=20.8$ J mol${}^{-1}$ K${}^{-1}$, so $C_p=\frac{7}{2}R=29.1$ J mol${}^{-1}$ K${}^{-1}$ and $\gamma =1.40$. Heating 2.0 mol by 10 K needs

$$2.0\times 20.8\times 10=416\text{ J at constant volume}2.0\times 29.1\times 10=582\text{ J at constant pressure}$$

The extra 166 J is the expansion work, $nR\Delta T=2.0\times 8.31\times 10$.

An everyday example. A sealed gas cylinder left in the sun warms at constant volume, so all the heat it absorbs goes into internal energy and raises its pressure.

The substance. Mayer's relation holds only for ideal gases — solids and liquids barely expand, so their two specific heats are almost equal.

The second law states that no engine can turn all the heat it absorbs into work, and that heat cannot flow by itself from a colder body to a hotter one, which limits engine efficiency and means refrigerators need work to pump heat.

Statements:

- Kelvin-Planck — no process can have as its only result the complete conversion of heat from a reservoir into work
- Clausius — no process can have as its only result the transfer of heat from a colder body to a hotter body

Heat engine. It absorbs heat $Q_1$ from a hot source, does work W and rejects $Q_2$ to a cold sink, with efficiency

$$\eta =\frac{W}{Q_1}=1-\frac{Q_2}{Q_1}{\eta }_{\text{Carnot}}=1-\frac{T_2}{T_1}$$

Refrigerator. It uses work W to remove heat $Q_2$ from the cold space and releases $Q_1=Q_2+W$, with coefficient of performance $\beta =\frac{Q_2}{W}$.

Worked example 1 — an engine. An engine absorbs 1000 J and rejects 700 J:

$$W=1000-700=300\text{ J}\eta =\frac{300}{1000}=0.30$$

An ideal engine between 500 K and 300 K could reach $1-\frac{300}{500}=0.40$.

Worked example 2 — a refrigerator. A fridge removes 600 J from its cabinet for every 150 J of electrical work, so $\beta =\frac{600}{150}=4.0$, and it releases $600+150=750$ J to the room.

An everyday example. The outdoor unit of a split air conditioner blows out hot air, because it releases the heat taken from the room plus the work done by the compressor.

The substance. A refrigerator does not break the second law — heat moves from cold to hot only because external work is supplied.