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Two moles of nitrogen are initially at 10 bar and 600 K (state 1) in a horizontal piston/cylinder device. They are expanded adiabatically to 1 bar (state 2). They are then heated at constant volume to 600 K (state 3). Finally, they are isothermally returned to state 1. Assume that N 2 is an ideal gas with a constant heat capacity as given on the back flap of the book. Neglect the heat capacity of the piston/cylinder device. Suppose that heat can be supplied or rejected as illustrated below. Assume each step of the process is reversible. Calculate the net work done overall.

Respuesta :

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!!

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