1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K. ![Fig. 6.4 — pressure-volume plot of the expansion: State (1) at 2.0 bar and 22.7 L, following the isotherm down to State (2) at 1.0 bar](/figures/thermodynamics/isothermal-expansion-two-states-pv-plot-exemplar-fig-6-4.svg)

1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the

NCERT Exemplar · Q49

Q. 1.0 mol of a monoatomic ideal gas is expanded from state

(1) to state

(2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state

(1) to state

(2) at 298 K.

Fig. 6.4 — pressure-volume plot of the expansion: State (1) at 2.0 bar and 22.7 L, following the isotherm down to State (2) at 1.0 bar — Figure 6.4

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Concept understanding — Irreversible Expansion Work

Irreversible Expansion Work – From Intuition to Precision

Imagine you have a gas trapped inside a cylinder with a piston. If you suddenly pull the piston outward, the gas expands rapidly into the newly available space. That's an irreversible expansion — the gas doesn't pass through a series of equilibrium states; it rushes, swirls, and settles only at the end.

Now, think about the work done by the gas during this process. Work, in physics, is force times displacement. For a piston, force is pressure times area, so work becomes $P \Delta V$ . But here's the catch: during an irreversible expansion, the pressure of the gas is not uniform throughout the cylinder. There are pressure gradients, turbulence, and the gas near the piston face may be at a different pressure than the gas deeper inside.

So how do we calculate the work done?

The Key Insight

The work done by the gas is determined by the external pressure it pushes against — not its own internal pressure. Why? Because the piston moves only in response to the net force acting on it. That net force comes from the external pressure on the other side of the piston.

Important

For any expansion (reversible or irreversible), the work done by the gas is:

$W = \int P_{\text{ext}} \, dV$

where $P_{\text{ext}}$ is the pressure exerted on the gas by the surroundings (the piston face).

During a reversible expansion, the gas is always in equilibrium with the surroundings, so $P_{\text{gas}} = P_{\text{ext}}$ at every instant. That's why you can replace $P_{\text{ext}}$ with $P_{\text{gas}}$ and integrate using the gas's equation of state.

During an irreversible expansion, $P_{\text{gas}}$ is not equal to $P_{\text{ext}}$ — and often, $P_{\text{ext}}$ is held constant (like when you suddenly release the piston against atmospheric pressure). In that case, the work simplifies dramatically:

W = P_{\text{ext}} \Delta V

The Intuitive Picture

Think of pushing a heavy box across a rough floor. The work you do depends on the force you apply (your "external" force), not on the internal stresses inside the box. Similarly, the gas does work against the external resistance it meets — the piston's opposing force.

If the external pressure is constant (say, 1 atm), the gas does work equal to $P_{\text{ext}} \times$ (change in volume), regardless of how chaotically it expands. The gas might have been at 10 atm initially, but it only does work against the 1 atm it actually pushes.

Watch out

A common mistake: using the gas's own pressure to calculate irreversible work. Unless the process is reversible, $P_{\text{gas}} \neq P_{\text{ext}}$ , and using $P_{\text{gas}}$ gives the wrong answer.

The Precise Statement

Irreversible expansion work is the work done by a gas when it expands through a series of non-equilibrium states. It is calculated using the external pressure that opposes the expansion:

W_{\text{irr}} = \int_{V_1}^{V_2} P_{\text{ext}} \, dV

For the most common case — expansion against a constant external pressure (like the atmosphere or a fixed weight on the piston): …

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For a reversible isothermal expansion of an ideal gas, the work done by the gas equals $nRT \ln(V_2/V_1)$ because temperature (and hence internal energy) stays constant. Here $w = -1718 \,\text{J}$ (negative because the gas does work on the surroundings).

Why this approach works

In an isothermal process the temperature remains fixed, so for an ideal gas the internal energy $U$ does not change ( $\Delta U = 0$ ). The first law then tells us that all the heat absorbed goes entirely into doing work: $q = -w$ .

For a reversible path the external pressure tracks the gas pressure infinitesimally closely at every instant, $p_{\text{ext}} = p = \frac{nRT}{V}$ . The work is then the integral of $p \, dV$ from the initial to the final volume, and because $T$ is constant we can pull $nRT$ out front and integrate $\frac{dV}{V}$ to get a logarithm. That logarithm can be written in terms of the volume ratio $V_2/V_1$ or equivalently the pressure ratio $p_1/p_2$ (since $pV = \text{const.}$ at fixed $T$ ).

Step-by-step calculation

1. Recognize the process type.

The expansion is both reversible and isothermal at $T = 298 \,\text{K}$ . We have $n = 1.0 \,\text{mol}$ , $p_1 = 2 \,\text{bar}$ , $p_2 = 1 \,\text{bar}$ , so $\frac{V_2}{V_1} = \frac{p_1}{p_2} = 2$ .

2. Write the work formula for reversible isothermal expansion.

The work done by the gas (our sign convention: work done by the system is negative) is

w = - \int_{V_1}^{V_2} p_{\text{ext}} \, dV = - \int_{V_1}^{V_2} \frac{nRT}{V} \, dV.

Because $T$ is constant,

w = -nRT \int_{V_1}^{V_2} \frac{dV}{V} = -nRT \ln\left(\frac{V_2}{V_1}\right).

w = -nRT \ln\left(\frac{V_2}{V_1}\right) = -nRT \ln\left(\frac{p_1}{p_2}\right).

3. Substitute the numbers.

Take $R = 8.314 \,\text{J mol}^{-1}\text{K}^{-1}$ , $n=1.0 \,\text{mol}$ , $T=298\,\text{K}$ , and $\ln(2) \approx 0.693$ .

$w = - (1.0)(8.314)(298) \ln(2) = - 2477.6 \times 0.693 \approx -1717 \,\text{J}.$ …

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Q.1.0 mol of a monoatomic ideal gas is expanded from state

Why this approach works

Step-by-step calculation

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