Question

Nitrogen (n2) gas within a piston–cylinder assembly undergoes a compression from p1 = 20 bar, v1 = 0.5 m3 to a state where v2 = 2.75 m3 . The relationship between pressure and volume during the process is pv1.35 = constant. For the n2 determine (a) the pressure at state 2, in bar, and (b) the work in kj.

Answer 1

Answer:

(a) 2.002 bar

(b) 1284 kJ

Explanation:

Given;

pv¹°³⁵ = constant

p v¹°³⁵ = k                         ------------------(a)

This implies that

p₁v₁¹°³⁵ = p₂v₂¹°³⁵           -----------------(i)

Where;

p₁ and v₁ are the state 1 pressure and volume of the gas respectively.

p₂ and v₂ are the state 2 pressure and volume of the gas respectively.

(a) From the question;

p₁ = 20bar

v₁ = 0.5m³

p₂ = unknown

v₂ = 2.75m³

Substitute these values into equation (i) as follows;

=> 20 x 0.5¹°³⁵ = p₂ x 2.75¹°³⁵

=> 20 x 0.3923 = p₂ x 3.9183

=> 7.846 = p₂ x 3.9183

Solve for p₂ ;

p₂ = 7.846 / 3.9183

p₂ = 2.002 bar

Therefore, the pressure at state 2, in bar, is 2.002

(b) The work done, W, is given by

W = $\int\limits^a _b {p} \, dv$             ----------------(ii)

Where;

a and b are the states 2 and 1 values of the volume of the gas. i.e

a = v₂ = 2.75m³

b = v₁ = 0.5m³

Calculate the value of k in equation (a) as follows;

p₁v₁¹°³⁵ = k

k = 20 x 0.5¹°³⁵

k = 7.846

From equation (a), we can then write;

=> p = 7.846 v⁻¹°³⁵

W = $\int\limits^a _b {7.846* v^{-1.35} } \, dv$

W = 7.846 x $\int\limits^a _b {v^{-1.35} } \, dv$

Substitute the limits a and b;

W = 7.846 x [ $\frac{v^{-0.35} }{-0.35}$]$|^{2.75}_{0.5}$

W = 7.846 [$\frac{2.75^{-0.35} }{-0.35}$ - $\frac{0.5^{-0.35} }{-0.35}$]

W = 7.846 [-2.005 - (-3.6416)]

W = 7.846 [-2.005 + 3.6416]

W = 7.846 [1.6366]

W = 12.84 bar.m³

1 bar = 10⁵N/m²

12.84 bar = 12.84 x 10⁵N/m²

=> 12.84 bar.m³ = 12.84 x 10⁵ Nm = 1284000Nm = 1284000J = 1284 kJ

Therefore, the workdone is 1284 kJ

Answer 2

Part a)

As we know that

$P_1V_1^{1.35} = P_2V_2^{1.35}$

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

$20* 0.5^{1.35} = P * (2.75)^{1.35}$

$P = 2 bar$

so final state pressure will be 2 bar

Part b)

now in order to find the work done

$W = \int PdV$

$W = \int \frac{c}{V^{1.35}}dV$

$W = c\frac{V^{-0.35}}{-0.35}$

$W = \frac{P_1V_1 - P_2V_2}{0.35}$

$W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J$