Numerical 1: Isothermal Work Done

Problem: 2 mol of an ideal gas expands isothermally at T = 300 K from $V_{1}$ = 2 L to $V_{2}$ = 6 L. Find Q, W, and $\Delta U$. (R = 8.314 J $mol^{-1}$ $K^{-1}$)

Step 1 — Process type: Isothermal → T = const → $\Delta U$ = 0.

Step 2 — Work done: $W = nRT\ln\!\left(\frac{V_2}{V_1}\right) = (2\ \text{mol})(8.314\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})\ln\!\left(\frac{6}{2}\right)$ $= (4988.4\ \text{J})\ln(3) = 4988.4 \times 1.0986\ \text{J} = \boxed{5480\ \text{J}}$

Step 3 — Heat absorbed: Since $\Delta U$ = 0, Q = W = 5480 J (heat flows in to sustain temperature during expansion).

Step 4 — Dimensional check: [n][R][T] = mol × J $mol^{-1}$ $K^{-1}$ × K = J ✓