Arun
Last Activity: 4 Years ago
V1=1l⇒ initial volume, V1=1×10−3m3V2=3l⇒ final volume, V2=3×10−3m3
At STP:
T=273K
P1=101.325kPa⇒ initial pressure
Y=1.4
Work done in adiabatic process, ΔW=−(Y−1P2V2−P1V1)
adiabatic ⇒PVγ=constant
⇒P1V1γ=P2V2γ
⇒P2=P1(V2V1)γ=1.01325(3×10−31×10−3)1.4
⇒P2=101.325×(3)1.41=(4.6555101.325)kPa
P2=21.7646KPa
ΔW=−(1.4−121.7646×103×3×10−3−101.325×103×1×10−3)
ΔW=−(0.465.2937−101.325)=90.5J
Considering only magnitude, the work done by the air will be 90.5J
