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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

1 mole of a monatomic ideal gas expands adiabatically as its temperature falls from 400 K to 300 K (R = 8.31). The work done by the gas is about:

Answer: 1250 J.

  • A ≈623 J
  • B ≈831 J
  • C ≈1250 J
  • D ≈2490 J

Correct answer: C. ≈1250 J

Explanation: Adiabatic: W = -&Delta;U = -nC<sub>v</sub>&Delta;T = -(3/2)(8.31)(300-400) &asymp; 1250 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 →