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

(a) 2.002 bar

(b) 1284 kJ

Step-by-step 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 [
`(v^(-0.35) )/(-0.35)`]
`|^(2.75)_(0.5)`

W = 7.846 [
`(2.75^(-0.35) )/(-0.35)` -
`(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 (c)/(V^(1.35))dV`

`W = c(V^(-0.35))/(-0.35)`

`W = (P_1V_1 - P_2V_2)/(0.35)`

`W = (20* 0.5 - 2 * 2.75)/(0.35)* 10^5 = 12.86 * 10^5 J`