2 mol of an ideal gas undergoes isothermal reversible expansion from 1 L to 10 L at 27 °C. Calculate the heat absorbed by the gas. (Take R = 8.314 J K⁻¹ mol⁻¹, ln 10 = 2.303.)

For an isothermal reversible change, $q = -w = nRT \ln(V_f / V_i)$. Substituting: $q = 2 \times 8.314 \times 300 \times \ln(10) = 2 \times 8.314 \times 300 \times 2.303 \approx 11488$ J ≈ **+11.49 kJ**. Heat is **absorbed** (positive q) because the gas does work on the surroundings while T is held constant. 5.74 kJ drops the factor of 2 (1 mol instead of 2). −11.49 kJ has the sign of w, not q. 49.88 kJ comes from using 10 instead of ln 10 (forgetting the logarithm).