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JEE Advanced Physics Thermodynamics 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Physics Question (2021) — Solution

Question

Paragraph A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, C_ V =2 R. Here, R is the gas constant. Initially, each side has a volume V_ 0 and temperature T_ 0 . The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V_ 0 / 2. Consequently, the gas temperatures on the left and the right sides become T_ L and T_ R , respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition. Question The value of  Q R T 0  is:

Options

  1. A. 4 2 2 + 1
  2. B. 4 2 2 - 1
  3. C. 5 2 + 1
  4. D. 5 2 - 1

Answer

B. 4 2 2 - 1

Step-by-step solution

Writing  C V  in terms of degree of freedom of the gas f , we get, C V = f 2 R = 2 R ⇒ f = 4 , Therefore,​ adiabatic index of gas will be given by,​​​​​​ γ = 1 + 2 f = 1 + 2 4 = 3 2 The gas of the right portion has undergone through an adiabatic process so, T i V i γ - 1 = T f V f γ - 1 ⇒ T 0   V 0 3 2 - 1 = T R V 0 2 3 2 - 1 ⇒ T R = 2 T 0 ⇒ T R T 0 = 2 Work done by the gas  2 ,  = - ∆ U = - n C V T f - T i = - 1 2 R 2 T 0 - T 0 = - 2 2 - 2 R T 0 So work done by the gas  1 , = + 2 2 - 2 R T 0 P V = n R T ⇒ P ∝ T V For gas  2 ,   T → 2  times of initial value, Volume  → 1 2  times of initial value So,  P ∝ 2 1 2 = 2 2  times of initial value ⇒ P f = 2 2 P 0 So pressure of the gas ( 1 ) will also be  2 2  times and volume of gas ( 1 ) becomes  3 2  times T ∝ P V ⇒ T ∝ 2 2 3 2 = 3 2  times of initial value T L = 3 2   T 0 ∆ U  for the left gas  = n C V T f - T i = 1 2 R 3 2 T 0 - T 0 = 6 2 - 2 R T 0 Heat given to gas ( 1 ) ,   Q = W + ∆ U = 2 2 - 2 R T 0 + 6 2 - 2 R T 0 = 4 RT 0 2 2 - 1 Therefore, Q R T 0 = 4 2 2 - 1

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