Respuesta :
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
