Thermodynamics — Medium JEE physics MCQ
**[JEE Mains 2026]** For an ideal gas, the volume is made 8 times and temperature is decreased to 1/4 of its initial value during an adiabatic process (q = 0). What is the value of γ (ratio of specific heats) for this gas?
- A. 5/3
- B. 7/5
- C. 4/3
- D. 3/2
Solution
For adiabatic process: TV^(γ-1) = constant.
T₁V₁^(γ-1) = T₂V₂^(γ-1).
T × V^(γ-1) = (T/4) × (8V)^(γ-1).
4 = 8^(γ-1) = 2^(3(γ-1)).
2² = 2^(3γ-3), so 2 = 3γ - 3, giving γ = 5/3.