For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the e

NCERT Exemplar · Q18

Q. For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w = -nRT ln(Vf/Vi). A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option. (Note: more than one of the given options may be correct.)

(i) Work done at 600 K is 20 times the work done at 300 K.

(ii) Work done at 300 K is twice the work done at 600 K.

(iii) Work done at 600 K is twice the work done at 300 K.

(iv) ΔU = 0 in both cases.

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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 an isothermal reversible expansion of an ideal gas, work is $w = -nRT \ln(V_f/V_i)$ and internal energy $\Delta U = 0$ because temperature is constant. Here, $V_f/V_i = 10$ , so $w \propto T$ . At 600 K, work is twice that at 300 K, and $\Delta U = 0$ in both cases.

Why this approach works

The First Law of Thermodynamics says $\Delta U = q + w$ . For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature doesn’t change, so $\Delta U = 0$ . That immediately tells us the heat absorbed equals the work done (with sign). The work formula for a reversible isothermal expansion is derived from $w = -\int P\,dV$ using the ideal gas law $P = nRT/V$ , giving $w = -nRT \ln(V_f/V_i)$ . Since $n$ and $R$ are constants, and the volume ratio is fixed, work is directly proportional to the absolute temperature $T$ .

Identify the given data
- $n = 1.0$ mol
- $V_f/V_i = 10$ (final volume is ten times initial)
- Two temperatures: $T_1 = 300$  K and $T_2 = 600$  K
- Process: isothermal and reversible expansion.
Write the work expression for each case

For the first experiment at 300 K:

w_{300} = - (1.0) R (300) \ln(10) = -300R \ln 10

For the second at 600 K:

w_{600} = - (1.0) R (600) \ln(10) = -600R \ln 10

Compare the magnitudes The negative sign indicates work is done by the system (expansion). Comparing absolute values:

\frac{|w_{600}|}{|w_{300}|} = \frac{600R \ln 10}{300R \ln 10} = 2

So work done at 600 K is twice that at 300 K. This matches option (iii).

Check the other options
- Option (i) says work at 600 K is 20 times that at 300 K — false, it’s 2 times. …

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