Thermodynamics ( solve the question)

Question

Calculate the amount of work required for adiabatic reversible expansion of 1 mole of

a polyatomic gas at –50°C and 10 atm pressure to 1 atm will be (given Cp/Cv = 1.33)?

Answer 1

We assume an ideal gas. We know p1 = 10 atm = 1.013 X 10⁴ Pa; p2 = 1 atm = 1.013 X 10⁵ Pa; T1 = -50 + 273 = 223 K; n = 1 mole; R = 8.31 J/(mol K) is the gas constant

We can use the Ideal Gas law to find V1, the initial volume.

V1 = nRT1 / p1 = (1 mole)(8.31 J/(mol K))(223 K) / (1.013 X 10⁴ Pa) = 0.1829 m³

For an adiabatic (no heat transfer) process, pV^γ = c, where c is a constant and in this case γ = 1.33.

Therefore, p1(V1)^γ = (1.013 X 10⁴)(0.1829)^1.33 = 1057.7 = c

We can use the same adiabatic relation to find V2.

p2 (V2)^γ = c

or

V2 - (c/p2)^1/γ = (1057.7/1.013 X 10⁵)^1/1.33 = 0.0324 m³

We can now calculate the work done in this adiabatic process.

W = ∫p dV from V1 to V2.

W = ∫ (c/V^γ) dV, substituting for the adiabatic relation.

W = c / (1 - γ) [V2^(1-γ) - V1^(1-γ)]

W = (1057)/(1 - 1.33) [0.0324^(1-1.33) - (0.1829)^(1-1.33)]

or

W = -(1057)/(0.33) [0.0324^(-0.33) - (0.1829)^(-0.33)]

W = -4325 J