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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

In an isochoric (constant volume) process, work done by the gas is:

Answer: Zero.

  • A Maximum
  • B Minimum
  • C Zero
  • D Equal to heat added

Correct answer: C. Zero

Explanation: W = P delta V. At constant volume, delta V = 0, so W = 0. All heat added goes to internal energy.

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 →