Chemical Thermodynamics Part 3

Concept:Work and energy changes for ideal gases depend on the process: reversible isothermal uses $W = -nRT\ln(V_2/V_1)$, irreversible isothermal uses $W = -P_{\text{ext}}\Delta V$, internal energy change is $\Delta U = nC_V\Delta T$, and enthalpy change is $\Delta H = nC_p\Delta T$.Explanation:For A (reversible isothermal expansion of 2 mol gas):$W = -nRT\ln(V_2/V_1) = -2 \times 8.314 \times 300 \times \ln(20/2) \ \text{J} = -11499 \ \text{J} \approx -11.5 \ \text{kJ}$.Magnitude is $11.5 \ \text{kJ}$, matches List-II entry II.For B (irreversible isothermal expansion of 1 mol gas against constant $P_{\text{ext}} = 3 \ \text{MPa}$? Actually 3 kPa, careful: $V_1=1 \ \text{m}^3$, $V_2=3 \ \text{m}^3$, $P_{\text{ext}}=3 \ \text{kPa}$ gives $W = -3 \times (3-1) \ \text{kJ} = -6 \ \text{kJ}$.Magnitude is $6 \ \text{kJ}$, matches entry III.For C (adiabatic expansion, $\Delta U$ depends only on temperature):$\Delta U = nC_V\Delta T = 1 \times \frac{3}{2}R \times 320 = 1.5 \times 8.314 \times 320/1000 \ \text{kJ} \approx 3.99 \ \text{kJ} \approx 4 \ \text{kJ}$.Matches entry I.For D (constant pressure, $\Delta H = nC_p\Delta T$):$\Delta H = 1 \times \frac{5}{2}R \times 337 = 2.5 \times 8.314 \times 337/1000 \ \text{kJ} = 7 \ \text{kJ}$.Matches entry IV.Answer:A-II, B-III, C-I, D-IV (Option D).