Question:

Select the correct match between List-I and List-II:
List-I:
(I) Isothermal reversible (1 mole ideal gas, T = 300K, 2dm$^3$ to 20dm$^3$) calculate $|w|$
(II) Isothermal irreversible [3KPa, 1m$^3$ to 3m$^3$] calculate $|w|$
(III) 1 mole gas undergoes constant pressure process in which change in temperature is 400K, C$_p$ = 5R/2, $\Delta H$ will be
(IV) 1 mole ideal gas having C$_v$ = 3R/2 and $\Delta T$ = 320K, calculate $\Delta U$
List-II:
(A) 8.32
(B) 6
(C) 4
(D) 5.74

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In thermodynamic processes, work done and energy changes can be calculated using the appropriate formulas, such as \(w = -nRT \ln \frac{V_2}{V_1}\) for isothermal processes and \(\Delta H = n C_p \Delta T\) for constant pressure processes.
Updated On: Jan 22, 2026
  • I-A: II-B: III-D: IV-C
  • I-D: II-B: III-A: IV-C
  • I-B: II-A: III-C: IV-D
  • I-A: II-B: III-C: IV-D
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The Correct Option is B

Solution and Explanation



Step 1: Analyzing each case.
(I) Isothermal reversible: For isothermal reversible expansion, the work done is calculated as: \[ w = -nRT \ln \frac{V_2}{V_1} \] Substituting the values: \[ w = -1 \times 8.31 \times 300 \times \ln \frac{20}{2} = -8.32 \, \text{kJ} \] Thus, the work done is 8.32 kJ. Therefore, the correct match is I-A. (II) Isothermal irreversible: For isothermal irreversible expansion, the work done is calculated as: \[ w = -P \Delta V \] Substituting the values: \[ w = -3 \times 10^3 \times (3 - 1) = -6 \, \text{kJ} \] Thus, the work done is 6 kJ. Therefore, the correct match is II-B. (III) Constant pressure process: For a constant pressure process, $\Delta H = n C_p \Delta T$. Given $\Delta T = 400$ K and $C_p = 5R/2$, the enthalpy change will be: \[ \Delta H = 1 \times \frac{5}{2} \times 8.31 \times 400 = 4 \times 10^3 \, \text{J} \] Thus, $\Delta H = 4 \, \text{kJ}$. Therefore, the correct match is III-A. (IV) Ideal gas with Cv: For an ideal gas, the change in internal energy is given by: \[ \Delta U = n C_v \Delta T \] Substituting the values: \[ \Delta U = 1 \times \frac{3}{2} \times 8.31 \times 320 = 5.74 \, \text{kJ} \] Thus, $\Delta U = 5.74 \, \text{kJ}$. Therefore, the correct match is IV-C.
Step 2: Conclusion. The correct matches are I-D, II-B, III-A, IV-C, which corresponds to option (2).
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