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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$g$
›Reveal solutionSolution
This tests irreversible isothermal expansion work against a constant external pressure, $w = -P_{ext}\Delta V$ , requiring volumes from the ideal gas law at both states. The work done by the gas is $-98.4\ J$ .

Concept and Intuition

When a gas expands against a constant external pressure (irreversible expansion), the work is simply $w = -P_{ext}(V_2-V_1)$ — not the reversible/isothermal-integral formula $w=-nRT\ln(V_2/V_1)$ , because the opposing pressure doesn't change as the gas expands. The negative sign reflects that the gas does work on the surroundings (loses energy).

Step-by-Step Solution

Moles of $O_2$ : $n = \dfrac{1.6}{32} = 0.05\ mol$ .

Initial volume (at $P_1=5\ atm$ , $T=300\ K$ ): $V_1 = \dfrac{nRT}{P_1} = \dfrac{0.05\times0.082\times300}{5} = \dfrac{1.23}{5} = 0.246\ L$ .

Final volume (after expanding until pressure equals the opposing $1\ atm$ , still at $T=300\ K$ since isothermal): $V_2 = \dfrac{nRT}{P_2} = \dfrac{1.23}{1} = 1.23\ L$ .

$\Delta V = V_2 - V_1 = 1.23-0.246 = 0.984\ L$ . …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$-163.3$
›Reveal solutionSolution
Using the initial pressure (1 atm, matching the external pressure) to fix $nRT=P_1V_1$ , the reversible isothermal work comes out to $-793.2\ J$ using $\log 5 = 0.7$ .

Concept and Intuition

For a reversible isothermal expansion of an ideal gas, the work done by the gas is $w = -nRT\ln\!\left(\dfrac{V_2}{V_1}\right) = -2.303\,nRT\log\!\left(\dfrac{V_2}{V_1}\right)$ . Here we aren't told $T$ directly, but the phrase "against an external pressure of 1 atm" tells us the gas starts in mechanical equilibrium with 1 atm at the initial volume — i.e. $P_1 = 1\,atm$ at $V_1=2\,L$ , so $nRT = P_1V_1$ (valid at constant $T$ , from the ideal gas law).

Step-by-Step Solution

$nRT = P_1V_1 = (1\ atm)(2\ L) = 2\ L\cdot atm$ (works in L·atm units directly, no need for R or T separately).

$w = -2.303\,nRT\log\!\left(\dfrac{V_2}{V_1}\right) = -2.303\times2\times\log\!\left(\dfrac{100}{2}\right) = -2.303\times2\times\log(50)$ .

$\log(50) = \log(5) + \log(10) = 0.7 + 1 = 1.7$ .

$w = -2.303\times2\times1.7 = -7.830\ L\cdot atm$ .

Convert to joules: $w = -7.830 \times 101.3\ J/(L\cdot atm) \approx -793.2\ J$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$J$
›Reveal solutionSolution
Gas evolved at constant (atmospheric) pressure does expansion work on the surroundings; $w=-\Delta n_g RT$ gives $-494.68$ J.

Concept and Intuition

In an open beaker the reaction occurs at constant (atmospheric) pressure, and any gas produced pushes back the atmosphere, doing $P$ - $V$ work. By the IUPAC convention $w=-P_{ext}\Delta V$ , work done by the system (expansion) is recorded as negative work done on the system. For an ideal gas at constant $T$ , $P\Delta V = \Delta n_{gas}RT$ .

Step-by-Step Solution

Reaction: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$ .

Moles of Zn $= \dfrac{13}{65} = 0.2$ mol.

From stoichiometry, moles of $H_2(g)$ produced $= 0.2$ mol, so $\Delta n_{gas} = 0.2$ (only the gas phase moles matter; solid Zn disappears, solute ions don't count). …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$q = -w = P_{ext}(V_{final} - V_{initial})$
›Reveal solutionSolution
This tests two standard first-law-of-thermodynamics results for isothermal irreversible and adiabatic processes. The answer is (A) -- both statements are correct.

Concept and Intuition

For an ideal gas, internal energy depends only on temperature. In an isothermal process, $\Delta T = 0$ so $\Delta U = 0$ , meaning all the heat absorbed exactly balances the work done ( $q = -w$ ). In an adiabatic process, no heat is exchanged with the surroundings ( $q=0$ ), so the entire change in internal energy shows up as work.

Step-by-Step Solution

Statement I: For an isothermal irreversible expansion of an ideal gas against a constant external pressure, $w = -P_{ext}(V_f - V_i)$ . Since $\Delta U = 0$ (isothermal, ideal gas) and $\Delta U = q + w$ , we get $q = -w = P_{ext}(V_f - V_i)$ . This exactly matches the given statement. Correct.

Statement II: For an adiabatic process, $q = 0$ by definition. The first law $\Delta U = q + w$ then reduces to $\Delta U = w_{adiabatic}$ . Correct. …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$B \rightarrow E$
›Reveal solutionSolution
This tests the standard graphical definition of thermodynamic work as the area under a P-V curve. The answer is (C).

Concept and Intuition

The work done during a process from one state to another on a P-V diagram equals the area enclosed between the process curve and the volume axis, over the volume range spanned by the process -- not any wider or narrower region. To find this area, drop perpendiculars (vertical lines) from the start and end points of the curve down to the V-axis; the enclosed region between those two verticals, the curve, and the V-axis segment between them is exactly the work done.

Step-by-Step Solution

The process of interest is from B to E along the given curve.

To find the work done for this specific process, drop a vertical line from B down to the V-axis (reaching point C) and a vertical line from E down to the V-axis (reaching point D).

The area representing the work done is bounded by: the curve from B to E (top), the vertical line E-D (right side), the V-axis segment D-C (bottom), and the vertical line C-B (left side) -- i.e., the region B to C to D to E. …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$Q. Which one of the following options is correct for q = -ve and w = +ve (A) Isothermal irreversible expansion of an ideal gas with constant external pressure. (B) Isothermal irreversible compression of an ideal gas with constant external pressure (C) Isothermal reversible expansion of an ideal gas against constant external pressure (D) Irreversible expansion of an ideal gas into vacuum$
›Reveal solutionSolution
For an isothermal ideal gas, $q=-w$ ; compression (work done ON the system, $w>0$ ) automatically means heat is released ( $q<0$ ) — this only happens for option (B).

Concept and Intuition

By the IUPAC sign convention: $w = -P_{ext}\Delta V$ , and work done on the system is positive. For an isothermal process on an ideal gas, internal energy doesn't change ( $\Delta U=0$ since $U$ depends only on $T$ ), so $q = \Delta U - w = -w$ . This means the signs of $q$ and $w$ are always exactly opposite in an isothermal ideal-gas process — so we just need to identify which process has $w>0$ (i.e., compression).

Step-by-Step Solution

Isothermal irreversible expansion (A): Gas expands, $\Delta V>0$ , so $w=-P_{ext}\Delta V<0$ (system does work, work is NOT done on it). Then $q=-w>0$ . Doesn't match (need $q<0$ ).

Isothermal irreversible compression (B): Gas is compressed, $\Delta V<0$ , so $w=-P_{ext}\Delta V>0$ (work done on system, e.g. by an external agent pushing the piston in). Then $q=-w<0$ (heat is released to keep $T$ constant while the gas is compressed). This matches $q=-ve, w=+ve$ .

Isothermal reversible expansion (C): Expansion again gives $w<0$ ; doesn't match. …
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AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$(1\ \mathrm{L.atm} = 101.32\ \mathrm{J})$
›Reveal solutionSolution
This tests the formula for work done during isothermal expansion against constant external pressure. Answer: X = 15 L.

Concept and Intuition

When a gas expands against a constant external pressure, the work done (using the convention $w = -P_{ext}\Delta V$ , work done ON the system, negative for expansion) depends only on the external pressure and the volume change — not on the path taken by the gas internally (unlike a reversible process).

Step-by-Step Solution

Convert given work to L·atm: $w = -2026.4\ \mathrm{J} \times \dfrac{1\ \mathrm{L\,atm}}{101.32\ \mathrm{J}} = -20.0$ L·atm.

Apply $w = -P_{ext}(V_2 - V_1)$ : $-20.0 = -2\,(X - 5)$ . …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$Q. 11.0 L of an ideal gas at constant external pressure of 5 atm is compressed isothermally to a final volume of one liter. The heat absorbed and work done respectively, during this compression (in L atm) are (A) -50, -50 (B) 50, -50 (C) -50, 50 (D) 50, 50$
›Reveal solutionSolution
Constant-external-pressure isothermal compression: work done on the gas is $+50$ L·atm, and since $\Delta U=0$ for an isothermal ideal gas, heat absorbed is $-50$ L·atm (i.e. 50 L·atm is released).

Concept and Intuition

This is an irreversible process (compression against a single constant external pressure, not a reversible quasi-static path), so work must be computed as $w=-P_{ext}\Delta V$ , not via the reversible integral $-nRT\ln(V_f/V_i)$ . Since the gas is ideal and the process is isothermal, internal energy doesn't change, so all the work done on the gas must leave as heat.

Step-by-Step Solution

Identify $P_{ext} = 5$ atm (constant, since it's an irreversible one-step compression), $V_i = 11$ L, $V_f = 1$ L.

Work done on the gas: $w = -P_{ext}(V_f - V_i) = -5\times(1-11) = -5\times(-10) = +50$ L·atm.

For an ideal gas at constant $T$ , $\Delta U = 0$ , so $q = \Delta U - w = 0 - 50 = -50$ L·atm. …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\rightleftharpoons$
›Reveal solutionSolution
A cyclic single-step (irreversible) isothermal process: compute $w=-P_{ext}\Delta V$ for each leg using the arrival-state pressure as the constant external pressure, then sum over the full round trip.

Concept and Intuition

All three states lie on the same isotherm ( $T=300$ K throughout, $n=1$ mol), so $V=RT/P$ at each state. A "single step" process moves the gas directly between two states against a CONSTANT external pressure (taken as the pressure of the state being moved into), so its work is $w=-P_{ext}\Delta V$ — this is NOT the same as the (zero) net work of a reversible cycle traced along a single isotherm, because each irreversible leg follows a straight P–V path rather than the curve, so forward and backward legs enclose a net nonzero area.

Step-by-Step Solution

Volumes: $V_1=\dfrac{RT}{15}$ , $V_2=\dfrac{RT}{10}$ , $V_3=\dfrac{RT}{5}$ , with $RT\approx25$ L bar (using $R\approx0.0833$ L·bar·mol $^{-1}$ K $^{-1}$ , $T=300$ K).

Cycle goes $1\to2\to3\to2\to1$ (out and back through the same states), each leg a single step against the pressure of the arriving state.

Leg $1\to2$ ( $P_{ext}=10$ ): $w_1=-P_{ext}(V_2-V_1)=RT\left(\dfrac{P_2}{P_1}-1\right)=RT\left(\dfrac{10}{15}-1\right)=-\dfrac{RT}3$ .

Leg $2\to3$ ( $P_{ext}=5$ ): $w_2=RT\left(\dfrac{P_3}{P_2}-1\right)=RT\left(\dfrac{5}{10}-1\right)=-\dfrac{RT}2$ .

Leg $3\to2$ ( $P_{ext}=10$ ): $w_3=RT\left(\dfrac{P_2}{P_3}-1\right)=RT\left(\dfrac{10}{5}-1\right)=RT$ . …
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AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
$20\ ^\circ C$
›Reveal solutionSolution
For an ideal gas heated at constant pressure, the work of expansion is simply $w=nR\Delta T$ . With $n=2$ and $\Delta T=20$ K, $w=40R$ .

Concept and Intuition

At constant pressure, work done during a volume change is $w = P\Delta V$ . Since $PV=nRT$ at both the initial and final states (constant $P$ , $n$ ), $P\Delta V = nR\Delta T$ — a clean shortcut that avoids needing $P$ or $V$ individually.

Step-by-Step Solution

Constant-pressure work: $w = P\Delta V$ .

From the ideal gas law at constant $P,n$ : $P\Delta V = nR\Delta T$ .

A temperature change of $20^\circ C$ is the same magnitude as $\Delta T = 20$ K (since the Celsius and Kelvin scales have equal-sized degrees). …
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AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$5\ m^3$
›Reveal solutionSolution
Irreversible expansion work against a constant external pressure is $P_{ext}\Delta V$ ; the work done ON the gas is $-50\ MJ$ .

Concept and Intuition

When a gas expands against a constant opposing (external) pressure, the work exchanged is $w = -P_{ext}\Delta V$ (work done ON the system, by the standard thermodynamic sign convention: expansion does negative work on the system since the system does positive work on surroundings). This does not require the reversible/integral formula since $P_{ext}$ is constant, not varying with volume.

Step-by-Step Solution

$\Delta V = V_2 - V_1 = 10 - 5 = 5\ m^3$ .

$P_{ext} = 10^7\ N.m^{-2}$ (constant, since it's an irreversible expansion against fixed external pressure).

Work done by the gas on the surroundings $= P_{ext}\Delta V = 10^7 \times 5 = 5\times 10^7\ J = 50\ MJ$ . …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$1\ L.bar = 100\ J$
›Reveal solutionSolution
Isoenthalpic + ideal gas ⇒ $\Delta T = 0$ ⇒ $\Delta U = 0$ ; apply the first law to the irreversible expansion work to get $Q = 10$ kJ.

Concept and Intuition

For an ideal gas, enthalpy and internal energy both depend only on temperature. So "isoenthalpic" ( $\Delta H = 0$ ) immediately forces $\Delta T = 0$ , which in turn forces $\Delta U = 0$ as well. With $\Delta U$ pinned to zero, the first law directly links the heat absorbed to the (irreversible, constant-external-pressure) work done.

Step-by-Step Solution

Ideal gas + isoenthalpic $\Rightarrow \Delta H = 0 \Rightarrow \Delta T = 0 \Rightarrow \Delta U = 0$ (since $U$ of an ideal gas depends only on $T$ ).

Irreversible expansion against constant $P_{ext}=10$ bar, $\Delta V = 30-20=10$ L: work done by the gas $= P_{ext}\Delta V = 10 \times 10 = 100$ L·bar $= 100\times100 = 10{,}000$ 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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