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

Work done in this process is given by

$W = \int P dV$

$W = \int (1.5V + 20)dV$

$W = 1.5\frac{V^2}{2} + 20V$

$W = 1.5\frac{5^2 - 1^2}{2} + 20*(5 - 1)$

$W = 98 *10^5 J$

so work done is given by above  equation