CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.


Solution

Let,
Initial pressure of the gas = P1
Initial volume of the gas = V1 
Final pressure of the gas= P2 
Final volume of the gas = V2 
Given, V2 = 2 V1, for each case.
In an isothermal expansion process,
work done=nRT1 lnV2V1
Adiabatic work done,
W=P1V1-P2V2γ-1
It is given that same work is done in both cases.
So,
nRT1 ln(V2V1)=P1V1-P2V2γ-1     ...(1)
In an adiabatic process,
P2=P1V1V2γ=P112γ
From eq (1),
nRT1 ln 2=P1V11-12γ×2γ-1
and nRT1 = P1V1
So, ln 2=1-12γ.2γ-1
Or (γ − 1) ln 2 = 1 − 21−γ

Physics
HC Verma - II
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image