Joule Expansion

Picture two rigid, thermally insulated bulbs joined by a stopcock. One bulb holds a gas at pressure $p$; the other is fully evacuated. The whole assembly sits in an adiabatic enclosure, immersed in a calorimeter that can detect tiny temperature shifts. Opening the stopcock lets the gas expand irreversibly into the vacuum until pressure equalizes. Two features make this process special. First, the gas expands against zero opposing pressure, so the work done by the gas is zero — not because the process is slow or reversible, but because there is nothing pushing back. Second, the enclosure is adiabatic, so no heat crosses the boundary. These two facts together, through $\Delta U = q + w$, pin down $\Delta U = 0$ regardless of what happens to the temperature during the violent, irreversible rush of gas. This is the crucial subtlety: the process is highly irreversible, with turbulence, pressure gradients, and non-equilibrium intermediate states. We cannot compute $q$ or $w$ along the actual path using equilibrium formulas. But because $U$ is a state function, we don't need to. We only need matching initial and final equilibrium states, and $\Delta U = 0$ tells us immediately that those two states share the same internal energy, whatever temperature difference (if any) separates them.

Because volume enters here as the natural control variable, and internal energy is held fixed by construction, the relevant response function is $(\partial T/\partial V)_U$: how much the temperature drifts per unit volume increase, at constant $U$. Contrast this with the more familiar constant-pressure Joule–Thomson coefficient, which is the subject of a related, more practically important experiment covered separately.

where $q=0$ from thermal insulation and $w=0$ from expansion against zero external pressure.

where $\mu_J$ is the Joule coefficient, units of K m⁻³ mol (or K/L per mole), measuring temperature response per unit volume change at fixed $U$.

where $C_V = (\partial U/\partial T)_V$ is the constant-volume heat capacity, linking the Joule coefficient to the internal pressure.

where the vanishing of the internal-pressure term is the defining thermodynamic signature of an ideal gas.

Example 1 — Confirming $\Delta T = 0$ for an ideal gas expansion (foundational). Suppose 2.00 mol of an ideal monatomic gas at 300.0 K occupies 10.0 L in one bulb, connected by a stopcock to an evacuated 15.0 L bulb, the whole assembly insulated. Find the final temperature and the entropy change. Temperature. Since the gas is ideal, $\mu_J = 0$ identically, so no matter how large the volume change,

We can cross-check via energy balance: $q=0$, $w=0$, so $\Delta U = 0$; and because $\Delta U = nC_{V,m}\Delta T$ for an ideal gas with $\Delta U=0$ and $C_{V,m}=\tfrac{3}{2}R \ne 0$, we must have $\Delta T = 0$. Consistent. Entropy change (a useful companion calculation). Even though $\Delta U = 0$ and $\Delta T = 0$, entropy still increases because volume increases irreversibly. Using the ideal-gas entropy formula for an isothermal volume change,

with $V_f = 10.0+15.0 = 25.0$ L and $V_i = 10.0$ L:

Sanity check: $\Delta S > 0$ as required for this spontaneous, irreversible process in an isolated system, confirming the Second Law even though temperature and internal energy are both unchanged.

Example 2 — Estimating the real-gas cooling from the Joule coefficient (applied, harder). For CO₂ gas near room temperature, experimental and van der Waals-based estimates give a Joule coefficient of approximately

To keep this concrete, use the van der Waals estimate directly. For a van der Waals gas, the internal pressure is

so the Joule coefficient becomes

For CO₂, $a = 0.3640\ \text{Pa m}^6\text{mol}^{-2}$ and $C_{V,m} \approx 28.5\ \text{J K}^{-1}\text{mol}^{-1}$ (roughly $\tfrac{7}{2}R$ for this near-room-temperature polyatomic gas including vibrational contributions). Take $n = 1.00$ mol expanding from $V_i = 5.00$ L $= 5.00\times10^{-3}\ \text{m}^3$ into a total final volume $V_f = 10.00$ L $=1.00\times10^{-2}\ \text{m}^3$ (doubling the volume by opening into an equal evacuated bulb). Estimate $\mu_J$ at the mean volume $\bar V \approx 7.5\times10^{-3}\ \text{m}^3$:

Now estimate the temperature drop over the finite volume change:

So doubling the volume of 1 mole of CO₂ by free expansion should cool it by roughly 1 K — small, but in principle measurable with sensitive calorimetry, unlike the strictly zero result for an ideal gas. Sanity check: the sign is negative because CO₂ molecules attract each other ($a>0$); pulling them apart against attraction costs energy that comes from kinetic energy, cooling the gas, exactly as physical intuition about intermolecular attraction predicts.

James Prescott Joule performed his free-expansion experiment in 1845, connecting two copper vessels — one containing air at about 22 atm, the other evacuated — submerged in a single water bath, and watching a thermometer for any sign of temperature change on opening the valve. He detected essentially nothing, and concluded (slightly too strongly, given his apparatus) that the internal energy of a gas depends on temperature alone. His calorimetry, though careful for its era, lacked the sensitivity to catch the roughly one-kelvin-scale cooling that real gases like CO₂ actually exhibit under comparable conditions, largely because his water bath thermally coupled to both vessels and diluted any localized temperature change. It fell to Joule himself, working later with William Thomson (Lord Kelvin), to design the much more sensitive porous-plug throttling experiment — the Joule–Thomson expansion — which successfully measured the small deviations from ideality that free expansion could only hint at.

Correction — the correct marked option should be the internal-pressure one:

Ask the tutor: How does the Joule–Thomson coefficient differ mathematically from the Joule coefficient, and why is throttling so much more sensitive to real-gas nonideality than free expansion?