An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alter

NCERT Exemplar · Q8

Q. An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alternatives. (Note: more than one of the given options may be correct.)

(a)

dU = 0

(b)

dQ = 0

(c)

dQ = dU

(d)

dQ = dW

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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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In an isothermal process the temperature stays constant, so internal energy doesn't change; all heat supplied goes into doing work. Options (A) and (D) are correct.

An isothermal process is one in which the temperature of the system remains constant throughout. For an ideal gas, this single constraint has powerful consequences that flow directly from the First Law of Thermodynamics.

The First Law states:

dQ = dU + dW

where $dQ$ is the heat added to the system, $dU$ is the change in internal energy, and $dW$ is the work done by the system.

For an ideal gas, internal energy depends only on temperature: $U = nC_VT$ (or more generally, $U = U(T)$ only). When temperature is held fixed, internal energy cannot change.

Let's examine each option:

Option (A): $dU = 0$

Since the process is isothermal, $\Delta T = 0$ . For an ideal gas, internal energy is a function of temperature alone. Therefore:

dU = nC_V \, dT = nC_V \cdot 0 = 0

This is correct.

Option (B): $dQ = 0$

This would describe an adiabatic process, not an isothermal one. In an isothermal expansion or compression, heat must be exchanged with the surroundings to maintain constant temperature. When the gas expands and does work, it would cool down unless heat flows in; conversely, compression would heat it unless heat flows out.

This is incorrect.
Option (C): $dQ = dU$ …

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

KEAM 2026Set eng-2026-04214 marksMCQ
$-11^{\circ}C$
›Reveal solutionSolution

Isochoric heating: $W=0$ , $Q=\Delta U=nC_V\Delta T=2\cdot C_V\cdot100=200\,C_V$ .

The process is at constant volume, so no work is done ( $W=0$ ) and all heat goes into internal energy. The temperature change is
$\Delta T=89-(-11)=100\ \text{K}.$
With $n=2$ mol and molar heat capacity $C_V$ ,

$Q=\Delta U=nC_V\Delta T=2\times C_V\times100=200\,C_V.$ …
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KEAM 2025Set eng-2025-04254 marksMCQ
$Q. If the heat supplied to an ideal gas is used up entirely to do work on the surrounding, then the process is called (A) isothermal compression (B) isothermal expansion (C) adiabatic compression (D) isobaric compression (E) isobaric expansion$
›Reveal solutionSolution

$Q$ entirely converted to work means $\Delta U = 0$ (constant $T$ ) with the gas doing positive work, i.e. an isothermal expansion.

First law of thermodynamics:
$Q = \Delta U + W.$
If the heat supplied is used entirely to do work on the surroundings, then $Q = W$ , which forces

$\Delta U = 0.$ …
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KEAM 2025Set eng-2025-04274 marksMCQ
$^{-1}$
›Reveal solutionSolution

For vaporization $\Delta n_g = +1$ , so $\Delta U = \Delta H - \Delta n_g RT = 41 - 3.1 = 37.9$ kJ mol $^{-1}$ .

For the process $\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(g)}$ , one mole of gas is produced, so $\Delta n_g = +1$ .

Using $\Delta H = \Delta U + \Delta n_g RT$ :
$\Delta U = \Delta H - \Delta n_g RT.$
At $T = 373\ \text{K}$ : …
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KEAM 2025Set eng-2025-04294 marksMCQ
$Q. Find the mismatch pair in the thermodynamic process (A) Isothermal : Absorption or emission of heat (B) Isobaric : Pressure constant (C) Isochoric : Volume constant (D) Irreversible : Loss of heat (E) Adiabatic : Heat exchange$
›Reveal solutionSolution
Adiabatic means $Q=0$ (no heat exchange), so the pairing in (E) is wrong.

Checking the pairs.

Isothermal — temperature constant, heat is absorbed or emitted: correct.

Isobaric — pressure constant: correct.

Isochoric — volume constant: correct.

Irreversible — accompanied by dissipative losses: acceptable. …
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KEAM 2024Set eng-2024-06054 marksMCQ
$Q. If an ideal gas, in an insulated vessel is allowed to expand into another similar evacuated vessel through a valve then (A) external work is done on the gas (B) the pressure of the gas is doubled (C) the volume of the gas is doubled (D) the pressure of the gas remains same (E) the temperature of the gas is increased$
›Reveal solutionSolution
In free (Joule) expansion into an identical empty vessel the gas fills twice the volume with $W=0$ , $Q=0$ , and $\Delta T=0$ for an ideal gas.

When the valve opens, the ideal gas expands into vacuum:

No external work is done ( $W=0$ , expansion against zero pressure).

The vessel is insulated ( $Q=0$ ), and for an ideal gas $U$ depends only on $T$ , so temperature is unchanged. …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$A$
›Reveal solutionSolution
In free expansion into vacuum, internal energy and temperature stay constant.

Concept and Intuition

When gas expands into an evacuated, thermally insulated vessel, it does no work (nothing to push against) and exchanges no heat. By the first law $\Delta U=Q-W=0$ , so $U$ is unchanged; for an ideal gas $U$ depends only on $T$ , so $T$ is unchanged.

Step-by-Step Solution

Insulated $\Rightarrow Q=0$ .

Expansion into vacuum $\Rightarrow W=0$ .

First law: $\Delta U=Q-W=0$ , so internal energy constant. …
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Q.An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alternatives. (Note: more than one of the given options may be correct.)

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