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ExamsJEE AdvancedPhysics

Two cylinders A (thermally insulated walls and piston) and B (maintained at temperature T0) have pistons connected by a massless rod. Both initially contain ideal gas (gamma = 1.5) at pressure P and temperature T0, each in volume V0. Cylinder A contains mass m of gas. The piston is slowly released and moves until cylinder A's gas volume becomes V0/2. What is the mass of gas in cylinder B?

  1. 2*sqrt(2) m
  2. 3*sqrt(2) m
  3. sqrt(2) m
  4. none

Correct answer: 3*sqrt(2) m

Solution

A compresses adiabatically to V0/2: PA' = P*2^gamma = P*2¹.5 = 2*sqrt(2)*P. B expands isothermally to 3*V0/2: PB' = P*V0/(3*V0/2) = 2P/3. At equilibrium PA' = PB': 2*sqrt(2)*P_A_initial = (2/3)*P_B_initial. Since initial pressures are equal, this gives a contradiction unless we use the mass relation: PB_initial/PA_initial = mB/m. Setting PA' = PB': 2*sqrt(2)*P = (2P/3)*(mB/m) is wrong framing. The correct approach: initial PA=PB=P (both same T and V, different masses means different pressures — wait, rereading: same T, same V, same P but different masses means same molar amounts... Let me re-read: 'cylinder A contains m gm at P and T0', 'cylinder B contains identical gas at T0 but different mass, volume V0, and we are told same initial pressure P'. So nA = PV0/RT0 = nB? Only if masses are same. The problem says 'different mass' so they may have different pressures initially — or the piston is held in place so they can have different pressures. At equilibrium: PA' = PB' (piston free). PA' = P*(2)¹.5 = 2*sqrt(2)*P (adiabatic for A). For B: isothermal, PB_initial*V0 = PB'*(3V0/2), so PB' = (2/3)*PB_initial. Setting equal: 2*sqrt(2)*P = (2/3)*PB_initial => PB_initial = 3*sqrt(2)*P. Since PB_initial = mB*R*T0/(M*V0) and P = m*R*T0/(M*V0): mB/m = PB_initial/P = 3*sqrt(2). So mB = 3*sqrt(2)*m.

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