Question

2.35 kg of oxygen as an ideal gas at 4.00 bar and 400 ° C develop an adiabatic process up to a pressure of 0.200 bar. What is the work in kJ, done during this process?

Answer 1

Mass of oxygen = 2.35 kg

Molar mass of oxygen = 16 g / mol = 0.016 kg / mol

We know that;

Moles = Mass / Molar mass

Therefore;

Moles of oxygen ; n = 2.35 / 0.016 = 146.875 mol

Temperature initially; T₁ = 400 °C = 673 K

Pressure initially ; P₁ = 4 bar

Pressure finally ; P₂ = 0.2 bar

We know that for an ideal gas undergoing adiabatic change:

T^r P^{(1 - r)} = constant

where r = C_p / C_v

For oxygen which is a diatomic gas;

C_v = 2.5 R

C_p = 3.5 R

r = 3.5 R / 2.5 R = 1.4

Therefore;

T^1.4 P^0.4 = constant

or

T₁^1.4 P₁^0.4 = T₂^1.4 P₂^0.4

Substituting the values:

(673)^1.4 (4)^0.4 = T₂^1.4 (0.2)^0.4

which gives;

T₂ = 673 X (4 / 0.2)^{(0.4 / 1.4)}

T₂ = 673 X0.286

T₂ = 192.5 K

From the first law of thermodynamics;

dU = dQ + W

For adiabatic process;

dQ = 0

Therefore;

W = dU

For an ideal gas;

dU = n C_v (T₂ - T₁)

W = n C_v (T₂ - T₁)

Substituting the values ;

W = 146.875 X 2.5 X 8.314 X (192.5 - 673)

W = - 1466869 J = -1467 kJ

This is the amount of work extracted from the system. Hence the sign is negative.