Question

An enclosed amount of nitrogen gas undergoes thermodynamic processes as follows: from an initial state A to a state B to C to D and back to A, as shown in the P-V diagram. Assume that the gas behaves ideally. (a) If the process A-B is isothermal, determine the pressure of the gas in the state B. (b) Calculate the heat transferred to the gas in the process A-B. (c) Calculate the work done on the gas, the heat transferred to the gas and the change in internal energy for the process B-C. (d) What is the change in internal energy of the gas for the entire process, A-B-C-D-A? point A(0.10,13.0) point B(0.12, P) C(0.135, 5.00) D(0.10, 5.00)........(V,P)

Answer 1

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;

$Q = V_1 \cdot p_1 \cdot ln \left (\dfrac{V_2}{V_1} \right) = 0.10 \times 13 \times ln \left (\dfrac{0.12}{0.10} \right) = 0.237 kJ = 237.02 J$

(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;

$Q = c_p \times p \times\dfrac{V_4 - V_3}{R} = \dfrac{5}{2} \times 4 \times (0.1 - 0.135) = -0.35 \ kJ$

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;

$Q = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ$

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.