Chemical Thermodynamics: Tricky Questions Solved (1)

Question

Calculate the work done when 2 moles of an ideal gas expand isothermally and reversibly from $1\text{ L}$ to $10\text{ L}$ at $300\text{ K}$. Take $R = 8.314\text{ J/(mol·K)}$.

Solution — Step by Step

Why This Works

For an isothermal process, internal energy $U$ doesn’t change ($\Delta U = 0$ for an ideal gas at constant $T$). By the first law, $\Delta U = q + w = 0$, so $q = -w = +11.49\text{ kJ}$. The gas absorbs heat from the surroundings exactly equal to the work it does.

The reversible path gives the maximum work an isothermal expansion can do. An irreversible expansion (e.g., free expansion into vacuum) does zero work. A finite-step irreversible expansion does intermediate work.

Alternative Method

Use $\log_{10}$ instead of $\ln$:

$w = -2.303\, nRT\log_{10}\frac{V_2}{V_1} = -2.303 \times 2 \times 8.314 \times 300 \times 1 \approx -11488\text{ J}$

Same answer. The factor $2.303$ converts natural log to base-10 log.

Common Mistake

Forgetting that $T$ stays constant in an isothermal process — students sometimes use $\Delta T \neq 0$ in mixed-up formulas. Also, plugging volumes in mL or m³ inconsistently. The ratio $V_2/V_1$ is dimensionless, so units cancel as long as both volumes are in the same unit.