Thermodynamics: Step-by-Step Worked Examples (10)

Question

An ideal gas undergoes an isothermal expansion at $T = 300\ \text{K}$ from volume $V_1 = 2\ \text{L}$ to $V_2 = 8\ \text{L}$. Find the heat absorbed and work done by $1$ mole of the gas. Take $R = 8.314\ \text{J/mol·K}$.

Solution — Step by Step

Final answer: $W = Q \approx 3458\ \text{J}$ (both positive — work done by gas, heat absorbed).

Why This Works

The first law $Q = \Delta U + W$ becomes a one-step problem the moment you identify the process. For ideal gases, internal energy depends only on temperature — so isothermal kills $\Delta U$ instantly.

The $\ln(V_2/V_1)$ form is the dead giveaway of an isothermal process. Memorise it alongside its sibling, the adiabatic relation $PV^\gamma = \text{const}$.

Alternative Method

Integrate $W = \int P\, dV$ with $P = \tfrac{nRT}{V}$. Same answer, but two extra lines of integration. Skip it in MCQs.

Common Mistake