Why Slow Expansion Gets More Work Out of a Gas Than a Sudden One

Burning LPG in a kitchen releases heat; a gas pushing a piston does work. Chemical thermodynamics tracks how energy flows in and out of a reacting substance, and whether heat, work or a change in stored energy is involved.

This part covers systems and properties, work in isothermal expansion, the first law, and enthalpy. Take $R=8.314$ J/mol K and $1$ L bar $=100$ J.

The system is the part being studied, the surroundings are everything else, and the boundary separates them; a system is open if it exchanges matter and energy, closed if it exchanges only energy, and isolated if it exchanges neither.

- Open: water boiling in an uncovered pan
- Closed: water heating in a tightly sealed steel flask
- Isolated: hot tea in an ideal vacuum flask

State and path functions. A state function — $U$, $H$, $p$, $V$, $T$ — depends only on the present state. A path function — heat $q$ and work $w$ — depends on how the change was carried out.

Extensive and intensive properties. Extensive properties depend on the amount of matter (mass, volume, internal energy, heat capacity); intensive ones do not (temperature, pressure, density, molar heat capacity).

Worked example. Pour together two $1$ L samples of water, each at $25$ °C. The volume becomes $2$ L — extensive — but the temperature stays $25$ °C — intensive.

An everyday example. Climbing to a hilltop temple by the steps or by the winding road gives the same gain in height — a state function — but very different distances walked — a path function.

The substance. Heat and work are not properties of a system; they exist only while energy is being transferred.

**With chemistry's convention — work done on the system is positive — expansion against a constant external pressure gives $w=-p_{ext}\Delta V$, reversible isothermal expansion of an ideal gas gives $w=-2.303nRT\log \frac{V_2}{V_1}$, and expansion into a vacuum does no work.

Signs.** $q>0$ when the system absorbs heat; $w>0$ when work is done on the system; expansion therefore gives negative $w$.

**Worked example — $2$ mol of ideal gas at $298$ K expands from $10$ L to $20$ L.

Irreversible, against $1$ bar:**

$$w=-1\times (20-10)=-10\text{ L bar}=-1000\text{ J}$$

Reversible:

$$w=-2.303\times 2\times 8.314\times 298\times \log 2\approx -3.43\times 10^3\text{ J}$$

Into a vacuum: $p_{ext}=0$, so $w=0$.

For an ideal gas at constant temperature $\Delta U=0$, so in each case $q=-w$.

An everyday example. Releasing the air from a tyre in one burst does little useful work, while letting it push a piston slowly could extract far more.

The substance. Reversible expansion gives the maximum work output, because the gas always pushes against almost its own pressure.

**The first law states that the change in internal energy of a system equals the heat added to it plus the work done on it, $\Delta U=q+w$ — energy is conserved.

Worked example 1.** A system absorbs $500$ J of heat and does $150$ J of work on its surroundings, so $w=-150$ J:

$$\Delta U=500+(-150)=+350\text{ J}$$

Worked example 2. $2.0$ kJ of work is done on a system while it gives out $0.5$ kJ of heat:

$$\Delta U=-0.5+2.0=+1.5\text{ kJ}$$

Worked example 3 — isothermal ideal gas. For the reversible expansion above, $\Delta U=0$, so the gas absorbs $q=+3.43\times 10^3$ J.

An everyday example. Heating milk in a closed container adds heat with almost no work, so nearly all of it raises the milk's internal energy.

The substance. **Physics often writes $\Delta U=Q-W$ with $W$ as work done by the system** — it is the same law with the opposite sign for work.

**Enthalpy is $H=U+pV$, so at constant pressure $\Delta H=\Delta U+p\Delta V$, which for reactions involving gases becomes $\Delta H=\Delta U+\Delta n_{g}RT$; for an ideal gas, $C_p-C_v=R$.**

Here $\Delta n_g$ is moles of gaseous products minus moles of gaseous reactants. $\Delta H$ is the heat absorbed at constant pressure, and $\Delta U$ the heat absorbed at constant volume.

Worked example 1 — ammonia. N${}_2$(g) + 3H${}_2$(g) $\to$ 2NH${}_3$(g), $\Delta H=-92.4$ kJ at $298$ K. With $\Delta n_g=2-4=-2$:

$$\Delta U=\Delta H-\Delta n_{g}RT=-92.4-(-2)(8.314\times 10^{-3})(298)\approx -87.4\text{ kJ}$$

Worked example 2 — vaporising water. H${}_2$O(l) $\to$ H${}_2$O(g) at $373$ K, $\Delta H=40.66$ kJ/mol, $\Delta n_g=1$:

$$\Delta U=40.66-8.314\times 10^{-3}\times 373\approx 37.56\text{ kJ/mol}$$

Heat capacities. $C_p={\left(\frac{\partial H}{\partial T}\right)}_p$ and $C_v={\left(\frac{\partial U}{\partial T}\right)}_V$; for one mole of an ideal gas, $C_p-C_v=R$.

An everyday example. Cooking in an open kadhai happens at constant pressure, so the heat involved is an enthalpy change.

The substance. **Only gases count in $\Delta n_g$** — liquids and solids change volume too little to matter.