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⚛️ Physics  ·  Thermodynamics  ·  NEET & JEE

An ideal gas undergoes free expansion into vacuum (Q=0, W=0). Temperature change:

Answer: Remains same.

  • A Increases
  • B Decreases
  • C Remains same
  • D Depends on gas

Correct answer: C. Remains same

Explanation: In free expansion, no work done (W=0) and no heat exchange (Q=0). So delta U = 0. For ideal gas, U depends on T only, so T stays constant.

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 →