JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given, sample of an ideal gas with (γ = 1.5) is taken through
adiabatic process, where
Initial volume, V1=1200cm3=12×10−4m3
Final volume, V2=300cm3=3×10−4m
Initial pressure, p1=200‌kPa=2×105×Pa
Let final pressure = p2
Using equation of adiabatic process,
p1V1γ=ρ2V2γ
p2=p1(

V1

V2

)γ=2×105(

12×10−4

3×10−4

)1.5
=16×105‌Pa
Work done in adiabatic process, W=

p1V1−p2V2

γ−1

Substituting the values, we get
W=

2×105×12×10−4−16×105×3×10−4

1.5−1

=−480J
Hence, work done in adiabatic compression is 480 J.