Share

## 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

Question

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.

in progress
0

Physics
2 weeks
2021-09-01T11:22:27+00:00
2021-09-01T11:22:27+00:00 1 Answers
0 views
0
## Answers ( )

Answer:

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