First law of thermodynamics — calculate work done in isothermal and adiabatic processes

Question

Using the first law of thermodynamics, derive expressions for work done in (a) an isothermal process and (b) an adiabatic process for an ideal gas. One mole of an ideal gas expands isothermally at 300 K from volume $V$ to $2V$ . Calculate the work done.

(NCERT Class 11, Chapter 12 — Thermodynamics)

Solution — Step by Step

The first law of thermodynamics:

\Delta Q = \Delta U + W

where $\Delta Q$ = heat supplied to the system, $\Delta U$ = change in internal energy, and $W$ = work done by the system.

For an ideal gas, $\Delta U = nC_v\Delta T$ .

In an isothermal process, temperature is constant ( $\Delta T = 0$ ). So $\Delta U = nC_v \times 0 = 0$ .

From the first law: $\Delta Q = W$ (all heat goes into doing work).

For work done by an ideal gas expanding from $V_1$ to $V_2$ :

W = \int_{V_1}^{V_2} P \, dV

Using $PV = nRT$ , so $P = nRT/V$ :

W = \int_{V_1}^{V_2} \frac{nRT}{V} dV = nRT \ln\frac{V_2}{V_1}

\boxed{W_{isothermal} = nRT \ln\frac{V_2}{V_1}}

In an adiabatic process, no heat exchange ( $\Delta Q = 0$ ). From the first law:

0 = \Delta U + W \implies W = -\Delta U = -nC_v(T_2 - T_1)

\boxed{W_{adiabatic} = \frac{nR(T_1 - T_2)}{\gamma - 1}}

where we used $C_v = R/(\gamma - 1)$ . Notice: the gas does positive work when it expands ( $T_1 > T_2$ , temperature drops).

$n = 1$ mol, $T = 300$ K, $V_1 = V$ , $V_2 = 2V$

W = nRT\ln\frac{V_2}{V_1} = 1 \times 8.314 \times 300 \times \ln 2

W = 2494.2 \times 0.693

\boxed{W \approx 1728 \text{ J} \approx 1.73 \text{ kJ}}

Why This Works

In an isothermal expansion, the gas maintains constant temperature by absorbing heat from the surroundings. Since temperature doesn’t change, internal energy stays the same, and all the absorbed heat is converted to work.

In an adiabatic expansion, no heat enters or leaves. The gas does work at the expense of its own internal energy — so it cools down. The temperature drop is directly proportional to the work done.

The key difference: isothermal work depends on the volume ratio ( $V_2/V_1$ ), while adiabatic work depends on the temperature change ( $T_1 - T_2$ ).

Alternative Method — Using PV relations for adiabatic work

For an adiabatic process, $PV^\gamma = \text{constant}$ . So:

W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{P_1 V_1^\gamma}{V^\gamma} dV = \frac{P_1V_1 - P_2V_2}{\gamma - 1}

This is equivalent to $\frac{nR(T_1 - T_2)}{\gamma - 1}$ since $PV = nRT$ .

For JEE and NEET, remember: isothermal work has a logarithm ( $nRT\ln$ ), adiabatic work has a fraction with $(\gamma - 1)$ in the denominator. If you see $\ln$ in the answer options, the process is likely isothermal. If you see $\gamma$ , it’s adiabatic.

Common Mistake

Students often confuse the sign convention. In the NCERT convention, $W$ is work done by the gas (positive for expansion). Some textbooks use work done on the gas (positive for compression). In NEET and JEE, the standard is work done by the gas. If the gas expands, $W > 0$ . If compressed, $W < 0$ . Always check which convention the question uses before writing your final answer.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using PV relations for adiabatic work

Common Mistake

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