Adiabatic Compression

220 kg of CO₂ gas at 27^oC and 1 atm is compressed adiabatically to 1/5th of its volume. It is then cooled to its original temperature at constant volume. Find Q, ΔU and W for each step and for the entire process.

Formula:

For an ideal gas, PV = nRT

For an adiabatic process, PV^γ = constant,

V₂/V₁ = (P₁/P₂)^(1/γ)

T₂/T₁ = (V₁/V₂)^(γ-1)

Adiabatic work of compression, W = (P₂V₂ - P₁V₁) / (γ - 1) → 1

Q + W = ΔU

ΔU = mC_V(T₂ - T₁) → 2

Here there are two steps taking place.

1. Adiabatic compression, and
2. Constant volume heat removal.

n = 220 / Molecular weight of CO₂ = 220/44 = 5 kmol

V₁ = 5 x 8314 x (273 + 27) / (1.01325 x 10⁵) = 123.08 m³.

V₂ = (1/5) x 123.08 = 24.616 m³.

γ = 1.3 (for CO₂: data)

T₂ = 300 x (5)^0.3 = 486.2 K

P₁/P₂ = (V₂/V₁)^γ

P₂ = 1 / (1/5)^γ

i.e., P₂ = 8.103 atm.

Adiabatic compression work W can be calculated from equn.1 or equn.2 (since ΔU = Q + W; and Q = 0 for an adiabatic process).

W = (8.103 x 1.01325 x 10⁵ x 24.616 - 1.01325 x 10⁵ x 123.08) / 0.3

= 25.7985 MJ

Since the temperature of the system is returned to its original state, the internal energy change of the total process is zero. Therefore, ΔU for the heat removal step is = -25.7985 MJ.

For the constant volume cooling step, W = 0, and ΔU = -25.7985 MJ. Therefore, Q for this step = -25.7985 MJ.

Summary: