Entropy of Ideal-Gas Expansion

Open a valve between a gas-filled flask and an evacuated one, and the gas rushes in to fill both. No work is done, no heat is exchanged, yet something irreversible has clearly happened. That "something" is captured quantitatively by entropy. This example matters because it is the cleanest possible illustration of the Second Law. It shows that entropy is a state function, computable along any convenient reversible path, even when the real process is wildly irreversible. It also cements the distinction between $q_{rev}/T$ and the entropy change itself. This article derives $\Delta S$ for isothermal and general expansions of an ideal gas. You will see the reversible isothermal case worked from first principles. Then you will see free (Joule) expansion compared directly against it. Finally you will see how both fit into a single general formula for any change in $T$ and $V$ .

Key takeaways

For an ideal gas, $\Delta S$ depends only on initial and final states, never on the path taken.
Reversible isothermal expansion: $\Delta S = nR\ln(V_2/V_1)$ , with $\Delta U = 0$ and $q_{rev} = w_{rev}$ .
Free expansion into vacuum has $q = w = \Delta U = 0$ , yet $\Delta S$ is identical to the reversible case.
The general formula $\Delta S = nC_{V,m}\ln(T_2/T_1) + nR\ln(V_2/V_1)$ handles any simultaneous $T$ – $V$ change.
Free expansion is irreversible: $\Delta S_{universe} = \Delta S_{sys} > 0$ , since $\Delta S_{surroundings} = 0$ .
Key idea: Entropy is a state function, so you always calculate $\Delta S$ using a reversible path, whether or not the real process was reversible.

The core insight

Because entropy is a state function, $\Delta S$ for any ideal-gas expansion (reversible or not) between the same initial and final $(T,V)$ states is always $nR\ln(V_2/V_1)$ at constant $T$ ; the difference between reversible and irreversible paths shows up not in $\Delta S_{sys}$ but in whether $\Delta S_{surroundings}$ manages to cancel it.

The physical picture

Picture one mole of ideal gas confined to volume $V_1$ inside a cylinder fitted with a frictionless piston, held in a heat bath at temperature $T$ . If the piston is withdrawn infinitely slowly, the gas pressure is always essentially equal to the external pressure. The gas does maximum possible work, and heat flows in reversibly from the bath to keep $T$ constant, since $\Delta U = 0$ for an ideal gas at constant $T$ . Now consider the opposite extreme: the same gas, same initial volume $V_1$ , but this time separated by a stopcock from an evacuated chamber of volume $V_2 - V_1$ . Open the stopcock. The gas expands freely against zero external pressure. No work is done on or by the surroundings, no heat crosses the boundary, and the temperature of an ideal gas is unchanged (Joule's experiment showed this directly, since $U$ depends only on $T$ for an ideal gas). Both processes start and end at the same $T$ , $V_1$ , $V_2$ . Because entropy is a state function, both processes must have the same $\Delta S_{sys}$ . What differs is the fate of the surroundings.

Figure

Reversible expansion versus free expansion between the same states

The second figure below shows the general relation as a surface: $\Delta S$ depends jointly on the temperature ratio and the volume ratio, and the isothermal expansion is just one slice through it.

Figure

ΔS as a function of volume ratio at fixed temperature

The core relations

\Delta S_{sys} = nR\ln\!\left(\frac{V_2}{V_1}\right) \quad (\text{isothermal, ideal gas})

where: $n$ is moles of gas, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $V_1$ , $V_2$ are initial and final volumes.

\Delta S_{sys} = nC_{V,m}\ln\!\left(\frac{T_2}{T_1}\right) + nR\ln\!\left(\frac{V_2}{V_1}\right) \quad (\text{general } T,V \text{ change})

where: $C_{V,m}$ is molar heat capacity at constant volume, assumed constant over the range.

\Delta S_{sys} = nC_{p,m}\ln\!\left(\frac{T_2}{T_1}\right) - nR\ln\!\left(\frac{p_2}{p_1}\right) \quad (\text{equivalent, in } T,p)

where: $C_{p,m} = C_{V,m} + R$ for an ideal gas.

\Delta S_{universe} = \Delta S_{sys} + \Delta S_{surr}, \qquad \Delta S_{surr} = -\frac{q_{rev,\,surr\,to\,sys}}{T}

where: for free expansion $\Delta S_{surr} = 0$ because $q=0$ for the isolated system, giving $\Delta S_{universe} > 0$ .

Worked examples

Derivation AReversible isothermal expansion of an ideal gas

Goal: find $\Delta S$ when $n$ mol of ideal gas expands reversibly and isothermally from $V_1$ to $V_2$ at temperature $T$ .

Step 1. Start from the thermodynamic definition of entropy. For any reversible path connecting the two states,

dS = \frac{dq_{rev}}{T}.

This is valid here because we are told the path is reversible; entropy itself is path-independent, but we need a reversible path to compute it, and we are lucky enough to already have one.

Step 2. Apply the First Law along this path.

dU = dq_{rev} + dw_{rev} = dq_{rev} - p\,dV.

For an ideal gas, $U$ depends only on $T$ (Joule's law), and $T$ is constant, so $dU = 0$ along this path. Hence

dq_{rev} = p\,dV.

Step 3. Substitute the ideal-gas equation of state. With $p = nRT/V$ ,

dq_{rev} = \frac{nRT}{V}\,dV.

Step 4. Divide by $T$ and integrate.

dS = \frac{dq_{rev}}{T} = \frac{nR}{V}\,dV \quad\Longrightarrow\quad \Delta S = nR\int_{V_1}^{V_2}\frac{dV}{V} = nR\ln\!\left(\frac{V_2}{V_1}\right).

Step 5. Recognize path-independence. Because $S$ is a state function, this expression is the entropy change for any process (reversible or not) connecting the same initial state $(T, V_1)$ to the same final state $(T, V_2)$ — including free expansion, where no reversible path exists physically but one can still be imagined to compute $\Delta S$ .

\boxed{\Delta S = nR\ln\!\left(\frac{V_2}{V_1}\right)}

Example 1 — Foundational: reversible isothermal expansion. A sample of 2.00 mol of ideal gas expands reversibly and isothermally at 298 K from 10.0 L to 30.0 L. Find $\Delta S_{sys}$ , $\Delta S_{surr}$ , $q_{rev}$ , $w_{rev}$ , and $\Delta S_{universe}$ . Step 1: Entropy change of the system.

\Delta S_{sys} = nR\ln\!\left(\frac{V_2}{V_1}\right) = (2.00\ \text{mol})(8.314\ \text{J K}^{-1}\text{mol}^{-1})\ln\!\left(\frac{30.0}{10.0}\right)

= (16.63\ \text{J K}^{-1})(1.0986) = 18.3\ \text{J K}^{-1}.

Step 2: Work done by the gas (reversible, isothermal).

w_{rev} = -nRT\ln\!\left(\frac{V_2}{V_1}\right) = -(2.00)(8.314)(298)\ln(3.00)

= -(4955\ \text{J})(1.0986) = -5443\ \text{J} = -5.44\ \text{kJ}.

(Negative sign: work is done by the gas on the surroundings.) Step 3: Heat absorbed by the gas. Since $\Delta U = 0$ (isothermal, ideal gas),

q_{rev} = -w_{rev} = +5.44\ \text{kJ}.

Step 4: Entropy change of surroundings. The surroundings lose 5.44 kJ of heat reversibly at 298 K:

\Delta S_{surr} = \frac{-q_{rev}}{T} = \frac{-5443\ \text{J}}{298\ \text{K}} = -18.3\ \text{J K}^{-1}.

Step 5: Total.

\Delta S_{universe} = 18.3 + (-18.3) = 0.

Sanity check: a reversible process must give exactly zero net entropy change for the universe. It does. ✓

Example 2 — Applied: free expansion versus reversible expansion, same states. The same 2.00 mol of gas at 298 K is instead allowed to expand freely (Joule expansion) from 10.0 L into an evacuated vessel, reaching a final volume of 30.0 L. Compare $\Delta S_{sys}$ , $\Delta S_{surr}$ , and $\Delta S_{universe}$ with Example 1, and then find $\Delta S$ if the gas simultaneously cools from 298 K to 280 K during a separate, non-isothermal reversible expansion to the same final volume, taking $C_{V,m} = \tfrac{3}{2}R$ (monatomic ideal gas). Step 1: Since the initial and final states $(T = 298\ \text{K}, V_1 = 10.0\ \text{L})$ and $(298\ \text{K}, 30.0\ \text{L})$ are identical to Example 1, and $S$ is a state function,

\Delta S_{sys} = 18.3\ \text{J K}^{-1} \quad (\text{unchanged}).

Step 2: For free expansion, the system is thermally and mechanically isolated during the process: $q = 0$ and $w = 0$ , so no heat crosses into the surroundings.

\Delta S_{surr} = 0.

Step 3: Total entropy change.

\Delta S_{universe} = 18.3 + 0 = +18.3\ \text{J K}^{-1} > 0.

This confirms the process is irreversible, consistent with the Second Law ( $\Delta S_{universe} \geq 0$ ). Step 4: Now the non-isothermal case. Using the general formula with $T_1 = 298$ K, $T_2 = 280$ K, $V_1=10.0$ L, $V_2 = 30.0$ L, $n=2.00$ mol, $C_{V,m} = 1.5(8.314) = 12.47\ \text{J K}^{-1}\text{mol}^{-1}$ :

\Delta S = nC_{V,m}\ln\!\left(\frac{T_2}{T_1}\right) + nR\ln\!\left(\frac{V_2}{V_1}\right)

= (2.00)(12.47)\ln\!\left(\frac{280}{298}\right) + 18.3\ \text{J K}^{-1}

= (24.94\ \text{J K}^{-1})(-0.0620) + 18.3\ \text{J K}^{-1} = -1.55 + 18.3 = 16.7\ \text{J K}^{-1}.

Sanity check: cooling lowers entropy (negative first term), expansion raises it (positive second term); since expansion dominates here, $\Delta S$ is still positive but smaller than the pure isothermal value — physically reasonable.

Limiting cases & common traps

Situation	What happens to ΔS_sys	What happens to ΔS_surr	ΔS_universe
Reversible isothermal expansion	$nR\ln(V_2/V_1) > 0$	$-nR\ln(V_2/V_1)$	0
Free (Joule) expansion, same states	$nR\ln(V_2/V_1) > 0$ (same!)	0	$> 0$
Reversible compression ( $V_2<V_1$ )	negative	positive	0
$V_2 = V_1$ (no volume change)	0 (isothermal term vanishes)	—	—
Adiabatic reversible expansion	0 (by definition of adiabatic reversible)	0	0

The chemist behind it

James Prescott Joule's expansion experiment, cooling an ideal-gas concept into concrete apparatus, showed that gas expanding into a vacuum causes no temperature change, which is exactly what makes $\Delta U = 0$ and $dq_{rev}=p\,dV$ valid in Step 2 of the derivation above. Rudolf Clausius later gave the concept its name and its rigorous mathematical footing through the inequality $dS \geq dq/T$ , unifying Joule's empirical observation with the abstract Second Law statement that entropy of an isolated system cannot decrease.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822–1888 · coined "entropy" and formulated the Second Law inequality dS ≥ dq/T that governs this very example

Ask the tutor: How does this same free-expansion logic extend to derive the entropy of mixing two different ideal gases?

Why is ΔS the same for reversible and free expansion if one process is "orderly" and the other is "chaotic"?

Entropy is a state function: it depends only on the initial and final thermodynamic states $(T,V)$ , not on the trajectory connecting them. Both processes here start and end at identical $T$ , $V_1$ , $V_2$ , so $\Delta S_{sys}$ must be identical. What differs is $\Delta S_{surroundings}$ : zero for free expansion (isolated, no heat exchanged) but exactly $-\Delta S_{sys}$ for the reversible path, keeping $\Delta S_{universe}=0$ only in the reversible case.

Does free expansion violate the First Law since ΔU = 0 but no work is done?

No violation at all. The First Law states $\Delta U = q + w$ . In free expansion both $q=0$ and $w=0$ independently, since there is no external pressure to push against and no thermal reservoir contact, so $\Delta U = 0$ trivially. This happens to coincide with $\Delta U=0$ from Joule's law (ideal gas, constant $T$ ), but the two zero results come from different physical reasons.

Can ΔS ever be negative for an ideal gas expansion?

Not for isothermal expansion into a larger volume: $\ln(V_2/V_1) > 0$ whenever $V_2>V_1$ , making $\Delta S_{sys}$ strictly positive. However, in the general formula, if the gas simultaneously cools enough (large negative $\ln(T_2/T_1)$ term), the net $\Delta S$ could in principle become negative even with expanding volume, though such a path could not occur spontaneously in an isolated system.

How does this connect to entropy of mixing?

Mixing two different ideal gases, each initially confined to its own volume, then allowed to occupy the combined volume, is mathematically two independent free expansions happening simultaneously. Each gas's partial pressure drops as it expands into the shared volume, and the total entropy of mixing sums two terms of the form $nR\ln(V_{final}/V_{initial})$ , one per gas. This is developed in detail in the treatment of entropy of mixing.

Why use R and not the Boltzmann constant k in these formulas?

$R$ and $k_B$ describe the same physics at different scales: $R = N_A k_B$ , where $N_A$ is Avogadro's number. Since these formulas are written per mole ( $n$ moles of gas), the molar gas constant $R$ is the natural choice. Working per molecule instead, you would use $k_B$ and replace $n$ with the number of molecules $N$ ; the statistical mechanical derivation via $S = k_B \ln W$ gives the identical numerical result.