StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Two cylinders A and B each contain the same amount of an ideal diatomic gas at 300 K. The piston of cylinder A is free to move (constant pressure), while the piston of cylinder B is fixed (constant volume). The same amount of heat Q is supplied to the gas in each cylinder. If the rise in temperature of the gas in cylinder A is 30 K, what is the rise in temperature of the gas in cylinder B?

  1. 30 K
  2. 42 K
  3. 70 K
  4. 35 K

Correct answer: 42 K

Solution

For the same heat Q supplied: Q = n*Cp*delta_T_A (cylinder A, constant pressure) and Q = n*Cv*delta_T_B (cylinder B, constant volume). Therefore n*Cp*delta_T_A = n*Cv*delta_T_B, giving delta_T_B = (Cp/Cv)*delta_T_A = gamma * delta_T_A. For a diatomic gas, gamma = 7/5. So delta_T_B = (7/5) * 30 = 42 K.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →