When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.
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;
(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;
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;
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.