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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

Two moles of a monatomic ideal gas are warmed from 300 K to 400 K at constant pressure (R = 8.31 J/mol K). The work done by the gas is about:

Answer: 1662 J.

  • A 3320 J
  • B 1662 J
  • C 831 J
  • D 2493 J

Correct answer: B. 1662 J

Explanation: At constant pressure W = nRΔT = 2 × 8.31 × 100 = 1662 J.

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.

Read the full Thermodynamics notes →