V1=11⇒ initial volume, V1=1×10−3m3 V2=31⇒ final volume, V2=3×10−3m3 At STP: T=273K P1=101.325kPa⇒ initial pressure Y=1.4 Work done in adiabatic process,∆W=−(
P2V2−P1V1
Y−1
) adiabatic ⇒PVγ= constant ⇒P1V1γ=P2V2γ ⇒P2=P1(
V1
V2
)γ=1.01325(
1×10−3
3×10−3
)1.4 ⇒P2=101.325×
1
(3)1.4
=(
101.325
4.6555
)kPa P2=21.7646KPa ∆W=−(
21.7646×103×3×10−3−101.325×103×1×10−3
1.4−1
) ∆W=−(
65.2937−101.325
0.4
)=90.5J Considering only magnitude, the work done by the air will be 90.5J