A gas initially at P₁=10⁵ Pa and V₁=10⁻³ m³ reaches P₂=(1/32)×10⁵ Pa and V₂=8×10⁻³ m³ through a quasi-static adiabatic process satisfying P³V⁵=constant. Another path reaches the same final state by an isobaric expansion at P₁, followed by an isochoric process at V₂. The heat supplied in this two-step path is approximately:
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Question

A gas initially at P₁=10⁵ Pa and V₁=10⁻³ m³ reaches P₂=(1/32)×10⁵ Pa and V₂=8×10⁻³ m³ through a quasi-static adiabatic process satisfying P³V⁵=constant. Another path reaches the same final state by an isobaric expansion at P₁, followed by an isochoric process at V₂. The heat supplied in this two-step path is approximately:

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G1 Physics · Accepted Answer

From P³V⁵=constant, γ=5/3, so for one mole of monoatomic gas Cᵥ=3R/2. The initial and final temperatures follow from PV=nRT. For the two-step path, add the isobaric work P₁(V₂−V₁) to the total internal-energy change. This gives Q≈588 J. Option C.