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Question

Physics Question on work done thermodynamics

The initial pressure and volume of a given mass of an ideal gas (withCpCv=γ),\left(with\frac{C_{p}}{C_{v}} =\gamma\right), taken in a cylinder fitted with a piston, are P0 _0 and V0_0 respectively. At this stage the gas has the same temperature as that of the surrounding medium which is T0_0. It is adiabatically compressed to a volume equal to v02.\frac{v_{0}}{2}. Subsequently the gas is allowed to come to thermal equilibrium with the surroundings. What is the heat released to the surroundings ?

A

0

B

(2γ11)P0V0γ1\left(2^{\gamma-1}-1\right)\frac{P_{0} V_{0}}{\gamma-1}

C

γP0V0in2\gamma P_{0}V_{0} in 2

D

P0V02(γ1)\frac{P_{0}V_{0}}{2\left(\gamma-1\right)}

Answer

(2γ11)P0V0γ1\left(2^{\gamma-1}-1\right)\frac{P_{0} V_{0}}{\gamma-1}

Explanation

Solution

T0V0γ1=T(V02)γ1T=T02γ1T_{0}V_{0}^{\gamma-1} =T\left(\frac{V_{0}}{2}\right)^{\gamma-1} \Rightarrow T =T_{0}2^{\gamma-1}
Aftercompression,weassumethepistontobefixed.\therefore After compression, we assume the piston to be fixed.
ΔQ=nCvΔT=nRγ1(T0T02γ1)=P0V0γ1(12γ1)\therefore\Delta Q =nC_{v}\Delta T =n\frac{R}{\gamma-1}\left(T_{0}-T_{0}2^{\gamma-1}\right) =\frac{P_{0}V_{0}}{\gamma-1}\left(1-2^{\gamma-1}\right)
heatreleased=P0V0γ1(2γ11)\therefore heat released =\frac{P_{0}V_{0}}{\gamma-1} \left(2^{\gamma-1}-1\right)