Question
If 2 mole of an ideal gas expand isothermally and reversibly at 27^ from 1 dm ^3 to 1\ m ^3 calculate work done? [R = 8.314\ J K ^ -1 mol ^ -1 ]
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (20 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -49.95 kJ
- B. -99.90 kJ
- C. -34.46 kJ
- D. -68.92 kJ
Answer
C. -34.46 kJ
Step-by-step solution
Given: n = 2 mol T = 27^ C = 300 K V_1 = 1 dm ^3 V_2 = 1 m ^3 = 1000 dm ^3 R = 8.314 J K ^ -1 mol ^ -1 The work done in a reversible isothermal expansion is given by: W = -2.303 nRT _ 10 ( V_2 V_1 ) Substituting the given values: W = -2.303 2 8.314 300 _ 10 ( 1000 1 ) W = -2.303 600 8.314 3 W = -34464.8 J W = -34.46 kJ Answer: -34.46 kJ
Related: Chemistry — Thermodynamics (C) · All PYQ Banks