JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2026
- Hard
- MCQ
A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same \(P, V, T\). Heating is started from left side until pressure changes to \(\dfrac{27P}{8}\). If initial volume of each compartment was \(9\) litres then the final volume in right-hand side compartment is __________ litres. (for this ideal gas \(C_P/C_V = 1.5\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
The right compartment is enclosed by adiabatic walls and an adiabatic piston, and no heat is supplied to it. Therefore, the gas in the right compartment undergoes an adiabatic compression.
Since the piston is frictionless, the pressure in the right compartment is always equal to the pressure in the left compartment. The final pressure in both compartments is \(P_f = \dfrac{27P}{8}\).
For the adiabatic process in the right compartment, the equation is:
\(P_i V_i^\gamma = P_f V_f^\gamma\)
Given \(P_i = P\), \(V_i = 9\) L, \(P_f = \dfrac{27P}{8}\), and \(\gamma = \dfrac{C_P}{C_V} = 1.5 = \dfrac{3}{2}\). Substituting these values:
\(P(9)^{3/2} = \left(\dfrac{27P}{8}\right) V_f^{3/2}\)
Since \(9^{3/2} = (3^2)^{3/2} = 3^3 = 27\), the equation becomes:
\(27P = \dfrac{27P}{8} V_f^{3/2}\)
Dividing both sides by \(27P\):
\(1 = \dfrac{1}{8} V_f^{3/2}\)
\(V_f^{3/2} = 8\)
Taking the power of \(2/3\) on both sides:
\(V_f = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4\) litres.
Answer: \(4\)
Since the piston is frictionless, the pressure in the right compartment is always equal to the pressure in the left compartment. The final pressure in both compartments is \(P_f = \dfrac{27P}{8}\).
For the adiabatic process in the right compartment, the equation is:
\(P_i V_i^\gamma = P_f V_f^\gamma\)
Given \(P_i = P\), \(V_i = 9\) L, \(P_f = \dfrac{27P}{8}\), and \(\gamma = \dfrac{C_P}{C_V} = 1.5 = \dfrac{3}{2}\). Substituting these values:
\(P(9)^{3/2} = \left(\dfrac{27P}{8}\right) V_f^{3/2}\)
Since \(9^{3/2} = (3^2)^{3/2} = 3^3 = 27\), the equation becomes:
\(27P = \dfrac{27P}{8} V_f^{3/2}\)
Dividing both sides by \(27P\):
\(1 = \dfrac{1}{8} V_f^{3/2}\)
\(V_f^{3/2} = 8\)
Taking the power of \(2/3\) on both sides:
\(V_f = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4\) litres.
Answer: \(4\)
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