Question

a non-flow quasi-equilibrium process occurs for which p=â’1.5v+20 bar , where v is volume in m3. what is work done when v changes from 1 to 5 m3?

Answer 1

The formula is :

Wd = ∫pdv

p = 101325 pa

= 101325 × ₁∫⁵1.5V + 20

= ₁⁵[1.5/2V² + 20V]

= (1.5/50 + 100) - (1.5/2 + 20)

= 79.28

= 79.28 × 101325 = 803.3046kJ

Answer 2

Formula for word done Wd =  integral of pdv,

p = 101325 pa, = 101325 × integral of 1 to 51.5V + 20,

=  integral of 1 to 5 [1.5/2V² + 20V], = (1.5/50 + 100) - (1.5/2 + 20),

= 79.28, = 79.28 × 101325 = 803.3046kJ.

So v changes from 1 to 5 m3 then word done is 803.3046kJ