8 June, 02:02

# Air contained in a piston-cylinder assembly, initially at 2 bar, 200 K, and a volume of 1 L, undergoes a process to a final state where the pressure is 8 bar and the volume is 2 L. During the process, the pressure-volume relationship is linear. Assuming the ideal gas model for the air, determine the work and heat transfer, each in kJ.

+3
1. 8 June, 02:35
0
W = 0.5 KJ

Q = 6.97 KJ

Explanation:

Given

P₁ = 2 bar = 200000 Pa

V₁ = 1 L = 0.001 m³

T₁ = 200 K

P₂ = 8 bar = 800000 Pa

V₂ = 2 L = 0.002 m³

Workdone = PdV

If the pressure volume relationship is linear, then

W = P (avg) (V₂ - V₁)

P (avg) = (P₁ + P₂) / 2 = (2+8) / 2 = 5 bar = 500000 Pa

W = 500000 (0.002 - 0.001) = 500 J = 0.5 KJ

b) To calculate the heat transfer, we will apply the mathematical statement of the first law of thermodynamics.

Q - W = ΔU

Q = ΔU + W

ΔU = m (C₂T₂ - C₁T₁)

We need to obtain T₂ first

Using the PVT relation for ideal gases

P₁V₁/T₁ = P₂V₂/T₂

T₂ = (P₂V₂T₁) / (P₁V₁) = (800000*0.002*200) / (200000*0.001)

T₂ = 2400 K

C₁ = specific heat capacity of air at the temperature T₁ = 200 K is 1.07 KJ/K. kg (from literature)

C₂ = specific heat capacity of air at the temperature T₂ = 2400 K is 0.8625 KJ/K. kg (from literature)

Then we calculate mass from the ideal gas relation, P₁V₁ = mRT₁

R = gas constant for air = 287 KJ/kg. K

(200000 * 0.001) = m * 287 * 200

m = 0.003484 kg

ΔU = m (C₂T₂ - C₁T₁)

ΔU = 0.003484 [ (0.8625) (2400) - (1.07) (200) ]

ΔU = 6.47 KJ

Q = ΔU + W = 6.47 + 0.5 = 6.97 KJ