JEE Main 2026 — Thermodynamics Question with Solution

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 $P_f=\frac{27P}{8}$.

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, $P_f=\frac{27P}{8}$, and $\gamma =\frac{C_P}{C_V}=1.5=\frac{3}{2}$. Substituting these values:
$P(9{)}^{3/2}=\left(\frac{27P}{8}\right)V_f^{3/2}$

Since $9^{3/2}=(3^2{)}^{3/2}=3^3=27$, the equation becomes:
$27P=\frac{27P}{8}{V}_f^{3/2}$

Dividing both sides by $27P$:
$1=\frac{1}{8}{V}_f^{3/2}$

$V_f^{3/2}=8$

Taking the power of $2/3$ on both sides:
$V_f=(8{)}^{2/3}=(2^3{)}^{2/3}=2^2=4$ litres.

Answer: $4$