Question

During the execution of a reversible nonflow process where the volume changes 

from an initial volume is 0.845 m3

to a final volume of 0.28 m2

and the pressure varies 

as p = -730V + 690 kPaa, where V is in m3

. Determine the work done

Answer 1

Solution;

Given;

$v_1=0.845m^3$

$v_2=0.28m^3$

$p=(-730v+690)kPa$

Now;

$W=\int_{v_1}^{v_2}pdv$

$W=\int_{0.845}^{0.28}(-730v+690)dv$

$W=[-365v^2+690v]_{0.845}^{0.28}$

$W=164.58-322.43=-157.85kJ$