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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

1 mole of diatomic gas (Cv = 5R/2) undergoes isochoric heating from 300K to 600K. Heat added:

Answer: 6235 J.

  • A 1247 J
  • B 4157 J
  • C 6235 J
  • D 8314 J

Correct answer: C. 6235 J

Explanation: Q = n Cv delta T = 1 x (5/2 x 8.314) x 300 = 5/2 x 8.314 x 300 = 6235.5 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 →