Answer:
W = 100.44 KJ
W(minimum) = 7.23KJ
Explanation:
Using Ideal gas
Given that ,
P₁ and P₂ = (100 and 135) Kpa respectively
V = 0.8m³
m = 1.54kg
R = 0.1889
T₁ = [tex]\frac{P_1V}{mR}[/tex]
= [tex]\frac{100 * 0.8}{1.54 * 0.1889}[/tex]
= 275k
T₂ = [tex]\frac{P_2V}{mR}[/tex]
= [tex]\frac{135 * 0.8}{1.54 * 0.1889}[/tex]
= 371k
The energies at initial and final state are determined for the given temperature using interpolation and were divided by molar mass of carbon dioxide
Given that
μ = internal energies
μ₁ = 141.56kjkg⁻¹
μ₂ = 206.78kjkg⁻¹
The actual work is the difference of the internal energies
W = m(μ₂ - μ₁)
= 1.54(206.78 - 141.56)kj
= 100.4kj
The zero entropies at the initial and final states are determined
∫ represent entropy
∫₁ = 4.788 kjkg⁻¹k⁻¹
∫₂ = 5.0478 kjkg⁻¹k⁻¹
(b)
The minimum work required is
w(minimum) = m(μ₂ - μ₁ - T₀(∫₂ - ∫₁))
= w - mT₀ (∫₂ - ∫₁ - RIn[tex]\frac{P_2}{P_1}[/tex]
= 100kj - 1.54 * 298 (5.0478 - 4.788 - 0.1889 * In[tex]\frac{35}{100}[/tex]) kj
= 7.23 kj
