What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{ext}$ in a single step as shown in Fig. 6.2. Explain graphically. ![Fig. 6.2 — an ideal gas in a cylinder compressed in a single step by a constant external pressure: the piston moves from the dashed initial level (volume V-I) to the final level (volume V-II)](/figures/thermodynamics/single-step-compression-constant-external-pressure-exemplar-fig-6-2.svg)

What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant externa

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p_{ext}

Fig. 6.2 — an ideal gas in a cylinder compressed in a single step by a constant external pressure: the piston moves from the dashed initial level (volume V-I) to the final level (volume V-II) — Figure 6.2

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Concept understanding — First Law of Thermodynamics

First Law of Thermodynamics

The Intuition: Energy is a Bank Account

Imagine you have a bank account. You can deposit money into it, or withdraw money from it. The total amount of money in your account changes only by the net of what goes in and what comes out. You cannot create money from nothing, nor can you destroy it — it just moves.

Energy works the same way. In any physical or chemical process, energy is never created or destroyed. It is only transferred from one place to another, or converted from one form to another. This is the First Law of Thermodynamics — the law of conservation of energy, applied to systems where heat and work are the currencies.

When you heat a gas in a piston, the gas expands and pushes the piston up. The energy you put in as heat doesn't vanish — part of it stays inside the gas (raising its temperature), and part of it leaves as work done on the piston. The total energy of the universe remains constant.

The Precise Statement

\Delta U = Q - W

Where:

$\Delta U$ = change in the internal energy of the system (the energy stored inside — kinetic energy of molecules, potential energy in bonds, etc.)
$Q$ = heat added to the system (positive if heat flows into the system)
$W$ = work done by the system on the surroundings (positive if the system expands and pushes against something)

The sign convention is crucial. Many textbooks use $Q + W$ with work done on the system. The version above ( $Q - W$ ) is the most common in Indian exam syllabi (CBSE, JEE, NEET). Stick to one convention and be consistent.

Watch out

Common Mistake

Students often forget the sign of work. If a gas expands, it does positive work on the surroundings — so $W$ is positive, and $\Delta U = Q - W$ becomes smaller. If a gas is compressed, work is done on the gas — so $W$ is negative, and $\Delta U = Q - (-|W|) = Q + |W|$ , which increases internal energy.

What Each Term Means Physically

Internal energy ( $U$ ) is the total microscopic energy of the system. For an ideal gas, it depends only on temperature — higher temperature means higher $U$ . For real substances, it also depends on volume and phase.

Heat ( $Q$ ) is energy transferred due to a temperature difference. If you put a hot pan on a cold stove, heat flows from pan to stove. In thermodynamics, we always ask: who is the system? If the system is the gas, then $Q$ is positive when heat flows into the gas.

Work ( $W$ ) in thermodynamics is usually pressure-volume work: $W = \int P \, dV$ . When a gas expands against a piston, it does work on the piston. When you compress a gas, you do work on it.

A Simple Example

Take a cylinder with a movable piston, containing 1 mole of an ideal gas. You supply 500 J of heat to the gas. The gas expands and does 200 J of work on the piston.

$Q = +500$ J (heat enters the system)
$W = +200$ J (work done by the system)

$\Delta U = 500 - 200 = 300 \text{ J}$ …

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For a single-step compression against constant external pressure, the work done on the gas is $W = p_{\text{ext}}(V_1 - V_2)$ , represented graphically by the area of a rectangle of height $p_{\text{ext}}$ and width $(V_1 - V_2)$ on a $p$ – $V$ diagram.

The First Law of Thermodynamics tells us that work is a path-dependent quantity — it depends on how the volume change happens, not just the start and end states. For a gas, the work done on the gas is given by

W = -\int_{V_1}^{V_2} p_{\text{gas}} \, dV

where $p_{\text{gas}}$ is the pressure of the gas at each instant. The negative sign ensures that when volume decreases ( $dV < 0$ ), work on the gas comes out positive.

But here’s the key: in a single-step compression against a constant external pressure $p_{\text{ext}}$ , the gas does not control the process — the surroundings do. The gas is forced to compress until its pressure equals $p_{\text{ext}}$ (or until the piston stops). During the entire compression, the external pressure remains fixed at $p_{\text{ext}}$ , so the work done on the gas by the surroundings is simply:

W = p_{\text{ext}} (V_1 - V_2)

Why no integral? Because $p_{\text{ext}}$ is constant, the integral becomes $p_{\text{ext}} \times (-\Delta V)$ , and since $V_2 < V_1$ , the result is positive — work is done on the gas.

Watch out

A common mistake is to write $W = p_{\text{ext}} (V_2 - V_1)$ . That gives a negative value for compression, which would mean work is done by the gas — the opposite of what happens. Always check the sign: compression → work done on gas → positive $W$ .

Now let’s walk through the reasoning step by step.

Identify the process. The gas is compressed from $V_1$ to $V_2$ in a single step. The external pressure $p_{\text{ext}}$ is constant throughout. This is an irreversible compression because the gas pressure is not equal to $p_{\text{ext}}$ during the process — it jumps suddenly.
Work formula for constant external pressure. When the external pressure is constant, the work done by the surroundings on the gas is:

W = p_{\text{ext}} \times (-\Delta V) = p_{\text{ext}} (V_1 - V_2)

Here $\Delta V = V_2 - V_1$ is negative, so $V_1 - V_2$ is positive.

Why not use gas pressure? In an irreversible process, the gas pressure is not uniform or well-defined during the compression — it may vary with position and time. The only pressure we can reliably use is the external one, because that’s what the surroundings actually apply. The formula $W = -\int p_{\text{gas}} dV$ is valid only if the gas is in equilibrium throughout (a reversible process). For a single-step irreversible compression, we must use $p_{\text{ext}}$ . …

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

Showing the 12 most recent of 19 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$\_\_\_\_$
›Reveal solutionSolution
For an ideal gas undergoing an isothermal process, temperature is constant, so internal energy (which depends only on temperature) does not change. By the first law of thermodynamics, any heat absorbed equals the work done by the gas. The correct option is (D).

The key concept here is the first law of thermodynamics combined with the properties of an ideal gas. The first law states:
$\Delta U = Q - W$
where $\Delta U$ is the change in internal energy, $Q$ is the heat added to the system, and $W$ is the work done by the system. For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature is constant, so $\Delta U = 0$ . This immediately tells us that $Q = W$ . The sign convention matters: if the gas expands, it does positive work and absorbs heat; if compressed, work is done on it and heat is released. The question’s phrasing “heat absorbed by the system is equal to the work done by the system” matches the expansion case, which is the standard interpretation.

Let’s walk through the reasoning step by step:

Identify the process: “Isothermal change” means the temperature of the gas remains constant throughout the process. For an ideal gas, internal energy is a function of temperature only (since intermolecular forces are negligible). Therefore, $\Delta U = 0$ .

Apply the first law: With $\Delta U = 0$ , the first law $\Delta U = Q - W$ becomes $0 = Q - W$ , so $Q = W$ . This means the heat added to the system equals the work done by the system. If the gas expands, $W > 0$ and $Q > 0$ (heat absorbed). If compressed, $W < 0$ and $Q < 0$ (heat released). The statement in option (D) says “heat absorbed by the system is equal to the work done by the system,” which is exactly the expansion scenario.

Evaluate the options:

(A) says “no change in internal energy” (correct) but “heat released by the system is equal to the work done by the system.” This would imply $Q$ negative and $W$ negative, which is possible for compression, but the phrasing “heat released” and “work done by the system” (positive work) are contradictory in sign. Typically, “work done by the system” is taken as positive, so this option is inconsistent.

(B) says internal energy increases — false for isothermal.

(C) says internal energy decreases — false. …
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COMEDK 2026Set 2026-M1 markMCQ
$100^{\circ} \mathrm{C}$
›Reveal solutionSolution
The change in internal energy is found from the first law: $\Delta U = Q - W$ , where $Q$ is the latent heat absorbed and $W$ is the work done against the atmosphere during expansion. The result is $21.01 \times 10^5 \, \mathrm{J}$ , which corresponds to option (D).

Concept & Intuition

When water boils at constant pressure (1 atm), it absorbs latent heat to break intermolecular bonds, but it also does work pushing the atmosphere back as it expands dramatically (from 1 cc to 1671 litres). The first law of thermodynamics tells us that the heat added ( $Q$ ) goes partly into increasing internal energy ( $\Delta U$ ) and partly into doing work ( $W$ ). So we compute $Q$ from the latent heat, $W$ from $P\Delta V$ , and subtract.

Step-by-step solution

Heat absorbed ( $Q$ ) The latent heat of vaporisation is $L = 22.68 \times 10^5 \, \mathrm{J/kg}$ . For 1 kg of water:

$Q = m L = 1 \times 22.68 \times 10^5 = 22.68 \times 10^5 \, \mathrm{J}.$

Work done ( $W$ ) At constant pressure, $W = P \Delta V$ .

Initial volume of water: $V_i = 1 \, \mathrm{cc} = 1 \times 10^{-6} \, \mathrm{m}^3$ .

Final volume of steam: $V_f = 1671 \times 10^3 \, \mathrm{cc} = 1671 \times 10^{-3} \, \mathrm{m}^3 = 1.671 \, \mathrm{m}^3$ .

Change in volume: $\Delta V = V_f - V_i \approx 1.671 \, \mathrm{m}^3$ (the tiny initial volume is negligible).

Pressure: $P = 1 \, \mathrm{atm} = 1.0 \times 10^5 \, \mathrm{Pa}$ . Hence, $W = P \Delta V = (1.0 \times 10^5) \times 1.671 = 1.671 \times 10^5 \, \mathrm{J}.$ …
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COMEDK 2025Set 2025-A1 markMCQ
$27^{\circ} \mathrm{C}$
›Reveal solutionSolution
The key idea is to relate enthalpy change ( $\Delta H$ ) to internal energy change ( $\Delta U$ ) via $\Delta H = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in moles of gas. For the reaction $\text{X(s)} + \frac12 \text{O}_2\text{(g)} \to \text{XO(s)}$ , $\Delta n_g = -0.5$ , giving $\Delta U = -601.83 - (-0.5)(8.314 \times 10^{-3})(300) = -600.58$ kJ/mol. The correct option is (D).

The relationship between enthalpy and internal energy is fundamental in thermochemistry. Enthalpy ( $H$ ) is defined as $H = U + PV$ , so for a constant-pressure process, the change in enthalpy equals the heat exchanged. The difference between $\Delta H$ and $\Delta U$ comes from the $PV$ work done when gases expand or contract. For reactions involving gases, the key formula is:
$\Delta H = \Delta U + \Delta n_g \, RT$
Here, the reaction is:
$\text{X(s)} + \frac12 \text{O}_2\text{(g)} \to \text{XO(s)}$
We are given $\Delta H = -601.83$ kJ/mol at $27^\circ$ C (300 K) and 1 bar. We need $\Delta U$ .

Step-by-step:

Determine $\Delta n_g$

Reactants: 1 solid + $\frac12$ mole of gas. Products: 1 solid (no gas).

So $\Delta n_g = 0 - \frac12 = -\frac12$ mol.

Convert units consistently

$R = 8.314$ J K $^{-1}$ mol $^{-1}$ = $8.314 \times 10^{-3}$ kJ K $^{-1}$ mol $^{-1}$ .

Temperature $T = 27 + 273 = 300$ K.

Compute the $PV$ work term

$\Delta n_g RT = (-0.5) \times (8.314 \times 10^{-3}) \times 300$

$= -0.5 \times 2.4942 = -1.2471$ kJ/mol.

Solve for $\Delta U$ …
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COMEDK 2024Set 2024-A1 markMCQ
$4.48 \mathrm{~L}$
›Reveal solutionSolution
Using the first law at constant volume, we find $C_v$ from the given heat and moles, then use $C_p = C_v + R$ to get $C_p = 6\ \text{cal mol}^{-1} \text{K}^{-1}$ , which corresponds to option (A).

Concept & Intuition

At constant volume, no work is done ( $W=0$ ), so the heat added equals the change in internal energy: $Q = n C_v \Delta T$ . Once we know $C_v$ , we can find $C_p$ using the relation $C_p - C_v = R$ for an ideal gas. The trick is to first find the number of moles from the given volume at STP.

Step-by-step solution

Find the number of moles of gas At STP (Standard Temperature and Pressure: $0^\circ\text{C}$ , 1 atm), 1 mole of an ideal gas occupies 22.4 L. Given volume = 4.48 L, so

$n = \frac{4.48}{22.4} = 0.2\ \text{moles}.$

Use the constant-volume heat equation At constant volume, $Q = n C_v \Delta T$ . Given $Q = 12.0$ cal, $\Delta T = 15^\circ\text{C} = 15\ \text{K}$ , and $n = 0.2$ :

$12 = 0.2 \times C_v \times 15$ $12 = 3 C_v \quad \Rightarrow \quad C_v = 4\ \text{cal mol}^{-1} \text{K}^{-1}.$

Apply the relation between $C_p$ and $C_v$ For an ideal gas: $C_p - C_v = R$ . Here $R = 2\ \text{cal mol}^{-1} \text{K}^{-1}$ (note: the problem gives $R$ in these units, so we use them directly).

$C_p = C_v + R = 4 + 2 = 6\ \text{cal mol}^{-1} \text{K}^{-1}.$
Watch out

A common mistake is to forget that $R$ is given in $\text{cal mol}^{-1} \text{K}^{-1}$ and to try converting units. The problem explicitly states $R=2\ \text{Cal kg}^{-1} \text{mol}^{-1}$ — but note the capital C in "Cal" is a typographical quirk; it means calories per mole per Kelvin. So just use $R=2$ directly.

Check the options …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{q}=0, \quad \Delta \mathrm{T} \neq 0, \quad \mathrm{w}=0$
›Reveal solutionSolution
For the free expansion of an ideal gas under adiabatic conditions, no heat is exchanged (q = 0), no work is done (w = 0), and because the internal energy of an ideal gas depends only on temperature, the temperature remains constant (ΔT = 0). The correct option is (D).

The key concept here is free expansion combined with adiabatic conditions. Free expansion means the gas expands into a vacuum — there is no opposing pressure, so the gas does no work. Adiabatic means no heat enters or leaves the system. For an ideal gas, internal energy depends only on temperature. Since no work is done and no heat is transferred, internal energy doesn’t change, so temperature stays constant.

Let’s walk through it step by step:

Adiabatic condition means the system is thermally insulated: no heat exchange with the surroundings. Therefore, $q = 0$ . This eliminates option (C), which claims $q \neq 0$ .

Free expansion means the gas expands into a vacuum. There is no external pressure to push against, so the gas does no work: $w = 0$ . This eliminates option (B), which claims $w \neq 0$ .

First Law of Thermodynamics states:

$\Delta U = q + w$
With $q = 0$ and $w = 0$ , we get $\Delta U = 0$ . …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{A}$
›Reveal solutionSolution

Internal energy is a state function, so $\Delta U_{B\to C} = 10$ J; adding the isobaric work $W_{B\to C} = 60$ J from the P–V diagram gives $Q_{B\to C} = 70$ J — option (C).

Because internal energy depends only on state,
$\Delta U_{B\to C} = U_C - U_B = 20 - 10 = 10\text{ J}.$
The step $B\to C$ is an isobaric expansion; the work done by the gas (area under $B\to C$ on the P–V diagram) is $W_{B\to C} = 60$ J. Applying the first law of thermodynamics, with $Q$ the heat supplied to the gas and $W$ the work done by the gas: …
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COMEDK 2024Set 2024-E1 markMCQ
$2240 \mathrm{~J}$
›Reveal solutionSolution

$\Delta U=Q-W=2240-168=2072$ J.

Heat supplied (latent heat for 1 g) $Q=2240$ J; work done by the vapour $W=168$ J.

First law of thermodynamics: …
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KCET 2023Set A-31 markMCQ
$8000\,\mathrm{J}$
›Reveal solutionSolution
The area enclosed by any closed $P$ – $V$ loop is the net work done; for a Carnot engine $W = \eta Q_H$ , and $\eta$ follows from the two isotherm temperatures read off the graph via $PV=nRT$ .

1. What the enclosed area means

For a cyclic process, $\Delta U = 0$ , so the first law gives $W_{\text{net}} = Q_{\text{net}}$ . Graphically, the area enclosed by the closed loop is exactly the net work done by the gas (positive for a clockwise loop, which this is). So the question is really: what is the work output of this Carnot engine?

2. Carnot efficiency
$\eta = \frac{W}{Q_H} = 1 - \frac{T_C}{T_H}$
We are told $Q_H = 8000\ \mathrm{J}$ , so we only need $T_C/T_H$ .

3. Get the temperatures from the graph

For $n = 1$ mole of an ideal gas, $T = \dfrac{PV}{R}$ , so temperature is proportional to the product $PV$ . On a Carnot cycle, $A\to B$ is the hot isotherm ( $T_H$ ) and $C\to D$ the cold isotherm ( $T_C$ ), so we may take the $PV$ product at any point on each isotherm. Use the two points whose coordinates fall exactly on printed gridlines:

Point $A$ (on the hot isotherm): $P = 1600\ \mathrm{kPa},\ V = 2.5\ \mathrm{cm^3}$

$(PV)_A = 1600 \times 2.5 = 4000 \ \ \mathrm{kPa\cdot cm^3}$

Point $C$ (on the cold isotherm): $P = 400\ \mathrm{kPa},\ V = 6.25\ \mathrm{cm^3}$ $(PV)_C = 400 \times 6.25 = 2500 \ \ \mathrm{kPa\cdot cm^3}$ …
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COMEDK 2023Set 2023-E1 markMCQ
$298 \mathrm{~K}$
›Reveal solutionSolution

Internal energy change: dU = dH - dn_g R T = -185.4 - (1.136)(8.314 x 10^-3 kJ/mol K)(298 K) = -185.4 - 2.81 = -188.2 kJ

Concept: dU = dH - (dn_g) R T, applied to the amount actually decomposed.

Reaction: 2 N2O(g) -> 2 N2(g) + O2(g); dH = -163.15 kJ (for 2 mol N2O)

dn_g = (2 + 1) - 2 = +1 (per 2 mol N2O)

Moles of N2O in 100 g (M = 44 g/mol):

n = 100/44 = 2.273 mol

Scale the reaction: factor = 2.273/2 = 1.136

dH = -163.15 x 1.136 = -185.4 kJ …
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COMEDK 2023Set 2023-M1 markMCQ
$q=0$
›Reveal solutionSolution

Adiabatic means “no heat transfer,” so $q=0$ .

By definition, in an adiabatic change the system is thermally insulated and exchanges no heat with the surroundings:

$q = 0.$ …
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COMEDK 2022Set 20221 markMCQ
$R$
›Reveal solutionSolution

Derivation: at constant volume, qv = Cv dT = dU. At constant pressure, qp = Cp dT = dH = dU + d(pV) = Cv dT + R dT for one mole of an ideal gas. Hence Cp = Cv + R, i.e. Cp - Cv = R.

Concept: Relation between molar heat capacities of an ideal gas (Mayer's relation).

Cp - Cv = R …
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COMEDK 2022Set 20221 markMCQ
$^\circ$
›Reveal solutionSolution

W = ηQ = 0.684 × 4.2 × 10⁵ ≈ 2.87 × 10⁵ J = 0.28 × 10⁶ J (≈0.29 × 10⁶).

Concept: Maximum (Carnot) efficiency η = 1 − T_sink/T_source, and W = ηQ.

T_sink = 27 °C = 300 K; T_source = 677 °C = 950 K.

η = 1 − 300/950 = 650/950 = 0.684.

Q = 100 kcal = 10⁵ cal = 10⁵ × 4.2 J = 4.2 × 10⁵ J. …
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Q.What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pextp_{ext}pext​ in a single step as shown in Fig. 6.2. Explain graphically.

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Q.What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{ext}$ in a single step as shown in Fig. 6.2. Explain graphically.

Unlock everything free for 14 days