Isothermal: W = nRT ln(V₂/V₁). Adiabatic: W = (P₁V₁ - P₂V₂)/(γ-1). Which does more work?

Work Done in Isothermal vs Adiabatic Expansion — Comparison

Question

A gas expands from volume $V_1$ to $V_2$ — once isothermally, once adiabatically. Both processes start at the same state $(P_1, V_1, T_1)$ . Which process does more work, and why?

This is a classic NCERT Class 11 Chapter 12 question that also appears regularly in JEE Main as a conceptual MCQ.

Solution — Step by Step

For an isothermal process (temperature constant, $T = T_1$ ):

W_{\text{iso}} = nRT_1 \ln\left(\frac{V_2}{V_1}\right)

For an adiabatic process (no heat exchange, $Q = 0$ ):

W_{\text{adi}} = \frac{P_1V_1 - P_2V_2}{\gamma - 1}

We need to compare these two values for the same $V_1 \to V_2$ expansion.

This is the key insight most students miss. Both start at $(P_1, V_1)$ , but as the gas expands, the pressure drops differently.

In isothermal expansion, temperature stays constant, so pressure falls slowly following $P = \frac{nRT_1}{V}$ .

In adiabatic expansion, the gas loses internal energy to do work (no heat comes in), so temperature also drops. Pressure falls faster — steeper curve on a $P$ - $V$ diagram.

Work done = area under the $P$ - $V$ curve. Draw both curves on the same graph starting at $(P_1, V_1)$ and ending at volume $V_2$ .

The isothermal curve sits above the adiabatic curve throughout the expansion. A higher curve = more area underneath = more work done.

W_{\text{iso}} > W_{\text{adi}}

For an adiabatic process, $PV^\gamma = \text{constant}$ where $\gamma > 1$ .

Since $\gamma > 1$ , pressure drops more steeply with volume compared to the isothermal case ( $PV = \text{constant}$ , equivalent to $PV^1$ ). The exponent 1 vs $\gamma$ ( $\gamma \approx 1.4$ for diatomic gases) explains why the adiabatic curve falls away faster.

For the same initial state and same volume change:

\boxed{W_{\text{isothermal}} > W_{\text{adiabatic}}}

The isothermal process always does more work during expansion.

Why This Works

The reason comes down to energy availability. In isothermal expansion, the system draws heat from the surroundings ( $Q = W$ , since $\Delta U = 0$ for an ideal gas at constant temperature). This heat continuously replenishes the energy being spent as work, keeping pressure higher.

In adiabatic expansion, there is no heat input. The only energy source is the gas’s own internal energy. As the gas does work, its temperature drops, and with it, the pressure drops faster. Less pressure means less force on the piston at each instant — so less work over the same volume change.

Think of it this way: the isothermal process gets an “external sponsor” (heat from surroundings). The adiabatic process is self-funded and runs out faster.

Alternative Method

We can also compare by expressing the adiabatic work in terms of $T_1$ and the final temperature $T_2$ .

Since $Q = 0$ , the first law gives $\Delta U = -W$ :

W_{\text{adi}} = nC_v(T_1 - T_2) = \frac{nR(T_1 - T_2)}{\gamma - 1}

Now for isothermal expansion to the same $V_2$ , the temperature stays $T_1$ .

But in the adiabatic case, $T_2 < T_1$ (the gas cools). If we were to expand isothermally to the same $V_2$ , the final pressure would be higher than in the adiabatic case. More pressure throughout = more work.

On a $P$ - $V$ diagram, if two curves start at the same point, the one with the smaller exponent in $PV^n = \text{const}$ lies higher. Isothermal has $n=1$ , adiabatic has $n=\gamma > 1$ . So isothermal always lies above adiabatic for the same starting point.

Common Mistake

Many students confuse “adiabatic does less work” with “adiabatic is less efficient.” They are different things. Adiabatic processes are actually used in heat engines specifically because no heat is wasted — but that’s a different question. For raw work output in expansion from $V_1$ to $V_2$ , isothermal always wins.

Another common error: using the isothermal formula $W = nRT\ln(V_2/V_1)$ for the adiabatic case by substituting a “final temperature.” This is wrong — the temperature changes continuously in an adiabatic process. You must use $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$ or $\frac{nR(T_1 - T_2)}{\gamma - 1}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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