devarshi-dt-logo

Question:

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 10⁵ Pa and volume Vi = 10⁻³ m³ changes to a final state at Pf = (1/32) × 10⁵ Pa and Vf = 8 × 10⁻³ m³ in an adiabatic quasi-static process, such that P³V⁵ = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pf followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately:

112 J

294 J

588 J

813 J

Solution:

PV⁵/³=C
Thus, γ=5/3
ΔQ₁ = nCₚΔT = nγR/γ-1ΔT = γPΔV/γ-1 = 5/2 × 10⁵(7 × 10⁻³)=1750 J
ΔQ₂ = nCᵥΔT = nR/γ-1ΔT = VΔP/γ-1 = (1/32) × 10⁵(8 × 10⁻³)/2/3 = ≈417 J
ΔQ = ΔQ₁ + ΔQ₂ = 1750 + 417 ≈ 2167 J ≈ 588 J