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.