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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

One mole of an ideal gas expands isothermally from V to 2V at 300 K (R = 8.31 J/mol·K). The work done by the gas is about:

Answer: 1.73 kJ.

  • A 0.69 kJ
  • B 1.73 kJ
  • C 3.46 kJ
  • D 8.31 kJ

Correct answer: B. 1.73 kJ

Explanation: W = nRT ln2 = 1·8.31·300·0.693 ≈ 1730 J ≈ 1.73 kJ.

Carnot Cycle (P-V Diagram)VP1→2: isothermal expansion (T_H)2→3: adiabatic expansion3→4: isothermal compression (T_C)4→1: adiabatic compression1234Net work done by the gas = area enclosed by the loop 1→2→3→4→1

The Carnot cycle alternates two isothermal steps (heat absorbed at TH, released at TC) with two adiabatic steps (no heat exchange); the enclosed loop area equals the net work output, and η = 1−TC/TH is the maximum possible efficiency between those two temperatures.

Concept context

Laws of thermodynamics, heat engines, entropy, and gas processes.

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