Answer:
Net work done overall = sum of work done for all the processes = 16,995.84 J
Explanation:
From the start, P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = ?
We can obtain V from PV = nRT; n = 2, R = 8.314 J/mol.K
V = 2 × 8.314 × 600/(1000000) = 0.009977 m³
P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = 0.009977 m³
For an adiabatic process for an ideal gas,
P(V^γ) = constant
γ = ratio of specific heats = Cp/CV = 1.4,
P₂ = 1 bar = 10⁵ Pa
P₁ (V₁^1.4) = P₂ (V₂^1.4) = k
10⁶ (0.009977^1.4) = 10⁵(V₂^1.4) = 1579.75 = k
V₂ = 0.0517 m³
Work done for an adiabatic process
W = k((V₂^(1-γ)) - (V₁^(1-γ))/(1-γ)
W = 1579.75 ((0.0517^0.4) - (0.009977^0.4))/0.4
W = 582.25 J
We still need T₂
PV = nRT
T₂ = P₂V₂/nR = 100000×0.0517/(2×8.314) = 310.92K
Step 2, constant volume heating,
Work done at constant volume is 0 J.
T₂ = 310.92K, T₃ = 600K
V₂ = 0.0517 m³, V₃ = V₂ = 0.0517 m³ (Constant volume)
P₂ = 1bar, P₃ = ?
PV = nRT
P₃ = nRT₃/V₃ = 2 × 8.314 × 600/0.0517 = 192974.85 Pa = 1.93bar
Step 3, isothermally returned to the initial state.
P₃ = 1.93bar, P₄ = P₁ = 10bar
T₃ = 600K, T₄ = T₁ = 600K (Isothermal process)
V₃ = 0.0517 m³, V₄ = V₁ = 0.009977 m³
Work done = nRT In (V₃/V₁) = 2 × 8.314 × 600 In (0.0517/0.009977) = 16413.59 J
Net work done = W₁₂ + W₂₃ + W₃₁ = 582.25 + 0 + 16413.59 = 16995.84 J
Hope this helps!!
