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7nadin3 [17]
3 years ago
12

A frictionless piston-cylinder contains carbon dioxide gas (CO2) initially at 500oC and 2 MPa. The system is o3 cooled in an iso

baric process until the final temperature becomes 350 C and the final volume is 1 m . For this process, determine: (a) The reduced pressure and the reduced temperature of the initial state. (b) The initial volume of the piston-cylinder (in m3). (c) The mass of CO2 in the piston-cylinder (in kg). (d) The total boundary work for the process (in kJ). (e) The amount of heat transfer during the cooling process (in kJ).
Physics
1 answer:
TiliK225 [7]3 years ago
8 0

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,

P_r = \dfrac{P}{P_c}

The critical pressure of carbon dioxide, P_c = 7.3773 MPa

P_r = \dfrac{2 \, MPa}{7.3773 \, MPa} \approx 0.2711 \, MPa

The reduced pressure, P_r = 0.2711 MPa

The critical temperature, T_c = 304.13 K

The reduced temperature, T_r, is given by the following formula;

T_r = \dfrac{T}{T_c}

Therefore, T_r = (773.15 K)/(304.13 K) = 2.54216947 K

The reduced temperature, T_r ≈ 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·(V_f - V_i)

W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227

The work done, W ≈ 388.023 kJ

(e) The heat transfer dQ = m·c_p ×(T₂ - T₁)

c_p 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

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