Respuesta :
Answer:
(a) 10.833 kPa
(b) 237.02 J
(c) The work done = -43.75 J
The heat transferred = 43.09 J
The change in internal energy = -0.6575 J
(d) The total change in internal energy for the entire process = 57.5 J
Explanation:
The parameters given are;
p₁ = 13 kPa
V₁ = 0.1 m³
p₂ = p
V₂ = 0.12 m³
p₃ = 5 kPa
V₃ = 0.135 m³
p₄ = 5 kPa
V₄ = 0.10 m³
(a) For isothermal process, we have
p₁V₁ = p₂V₂
p₂ = p₁V₁/V₂
∴ p₂ = 13 × 0.1/0.12 = 10.833 kPa
(b) The heat transfer for a constant temperature process is given by the relation;
[tex]Q = V_1 \cdot p_1 \cdot ln \left (\dfrac{V_2}{V_1} \right) = 0.10 \times 13 \times ln \left (\dfrac{0.12}{0.10} \right) = 0.237 kJ = 237.02 J[/tex]
(c) For process BC, we have
The work done is given by the area under the curve which between B and C which is given as follows;
0.5 × (V₃ - V₂) × (p₃ - p₂) = 0.5 * (0.135 - 0.12) * (5 -10.833) = -0.0437475 kJ = -43.75 J
For ideal gas, γ = 1.4
The heat transferred Q for a polytropic process is given by the relation
Q = W×(n - γ)/(γ - 1)
W = (V₂ × p₂ - V₃×p₃)(n - 1)
0.0427475 = (10.833 *0.12 - 5*0.135)/(n - 1)
(n - 1) = (10.833 *0.12 - 5*0.135)/0.0437475 =
n = 1 + 0.00575
n = 1.006
Q = -43.75*(1.006 - 1.4)/(1.4 - 1) = 43.09 J
The change in internal energy, u₃ - u₂ = Q + W = 43.09 J - 43.75 J = -0.6575 J
(d) For the constant pressure process DC, we have;
[tex]Q = c_p \times p \times\dfrac{V_4 - V_3}{R} = \dfrac{5}{2} \times 4 \times (0.1 - 0.135) = -0.35 \ kJ[/tex]
W = p×(v₄ - v₃) = 4*(0.1 - 0.134) = -0.136 kJ
u₄ - u₃ = Q + W = -0.35 - 0.135 = -0.485 kJ
For the constant volume process DA, we have;
[tex]Q = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]
Q = u₁ - u₄ = 1.2 kJ
Since process 1 to 2 is a constant temperature process we have u₁ - u₂ = 0
The total change in internal energy = u₁ - u₂ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄
The total change in internal energy = 0 - 0.6575 - 0.485 + 1.2 = 0.0575 kJ = 57.5 J
The total change in internal energy for the entire process = 57.5 J.
