A horizontal cylinder with adiabatic walls is closed at both ends and is divided into two volumes by a frictionless piston that is also thermally insulating. Initially the volume, pressure and temperature of the ideal gas in each side of the cylinder are equal to V0, P0 and T0 respectively. A heating coil in right side volume is used to heat slowly the gas on the side until the pressure reaches . (γ = 1.5)

Engineering

Physics

Thermodynamics

Question

A horizontal cylinder with adiabatic walls is closed at both ends and is divided into two volumes by a frictionless piston that is also thermally insulating. Initially the volume, pressure and temperature of the ideal gas in each side of the cylinder are equal to V₀, P₀ and T₀ respectively. A heating coil in right side volume is used to heat slowly the gas on the side until the pressure reaches $\frac{64 P_{0}}{27}$ . (γ = 1.5)

The pressure in LHS column is $\frac{4 P_{0}}{27}$

The volume in RHS column is $\frac{23 V_{0}}{16}$

The temperature in RHS column is $\frac{92 T_{0}}{27}$

The temperature in LHS column is $\frac{4 T_{0}}{3}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

New pressure will be same in both side for new temp. in left side

$P_{.}^{1 - γ} T_{0}^{γ} = {(\frac{64 P_{0}}{27})}^{1 - γ} T^{γ}$

$T = {(\frac{27}{64})}^{\frac{1 - γ}{γ}} T_{0} = {(\frac{27}{64})}^{\frac{1 - 1 .5}{1 .5}} T_{0}$

$= {(\frac{27}{64})}^{- \frac{1}{3}} T_{0} = \frac{4 T_{0}}{3}$

$\frac{64 P_{0} V^{1 .5}}{27} = P_{0} V_{0}^{1 .5}$

$V = V_{0} \times \frac{9}{16}$

$T = \frac{92}{27} T_{0}$