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Asked in JEE Main 7th Jan 1st Shift 2020 · Adiabatic Work and Internal Energy
Adiabatic work: W=(P₁V₁-P₂V₂)/(γ-1), γ-1=0.4.
At STP P₁=1.013×10⁵ Pa and V₁=10⁻³ m³, so P₁V₁=101.3 J.
P₂=(P₁)/(3^1.4)=(P₁)/(4.6555) and V₂=3×10⁻³ m³, so
P₂V₂=(3P₁V₁)/(4.6555)=65.28 J.
W=(101.3-65.28)/(0.4)=(36.02)/(0.4)=90.06 J.
The printed option is 90.5 J, which is 90.06 rounded loosely.
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