Respuesta :
Answer:
(a) The reduced pressure is 0.2711 MPa
The reduced temperature is 2.54 K
(b) The initial volume of the piston is approximately 0.806 m³
(c) The mass of CO₂ is approximately 16.9884 kg
(d) The work done, W is approximately 388.023 kJ
(e) The heat transfer is approximately -2,650.1904 kJ
Explanation:
The initial temperature of the piston-cylinder, T₁ = 500°C = 773.15 K
The initial pressure of the gas, P₁ = 2 MPa
The final temperature of the gas, T₂ = 350°C
The final volume of the gas = 1 m³
(a) For an isobaric process, we have;
The reduced pressure,
[tex]P_r = \dfrac{P}{P_c}[/tex]
The critical pressure of carbon dioxide, [tex]P_c[/tex] = 7.3773 MPa
[tex]P_r = \dfrac{2 \, MPa}{7.3773 \, MPa} \approx 0.2711 \, MPa[/tex]
The reduced pressure, [tex]P_r[/tex] = 0.2711 MPa
The critical temperature, [tex]T_c[/tex] = 304.13 K
The reduced temperature, [tex]T_r[/tex], is given by the following formula;
[tex]T_r = \dfrac{T}{T_c}[/tex]
Therefore, [tex]T_r[/tex] = (773.15 K)/(304.13 K) = 2.54216947 K
The reduced temperature, [tex]T_r[/tex] ≈ 2.54 K
(b) The initial volume of the piston, V₁ = (V₂/T₂) × T₁
∴ V₁ = (1 m³/773.15) × 623.15 = 0.80598848865 m³ ≈ 0.806 m³
The initial volume of the piston, V₁ ≈ 0.806 m³
(c) The number of moles of CO₂ in the cylinder, 'n', is given according to the following formula;
n = P·V/(T·R)
The universal gas constant, n = (2 × 10⁶Pa × 1 m³)/(623.15 K × 8.3145 J/(mol·K)) ≈ 386.0124 moles
The mass of CO₂ ≈ 386.0124 moles × 44.01 g/mol = 16.9884 kg
(d) The work done, W = P·([tex]V_f - V_i[/tex])
W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227
The work done, W ≈ 388.023 kJ
(e) The heat transfer dQ = m·[tex]c_p[/tex] ×(T₂ - T₁)
[tex]c_p[/tex] for CO₂ ≈ 1.04 kJ/(kg·K)
∴ dQ = 16.9884 × 1.04 × (350 - 500) = -2,650.1904 kJ
Therefore, the heat transfer = dQ = -2,650.1904 kJ
