Answer to Question #237530 in Molecular Physics | Thermodynamics for Randal Rodriguez

Question

3. During the execution of a reversible non-flow process the work is -156.2 KJ if V 1 =0.845 and the

pressure varies as P=-730 V +690 KPa, where volume V is in m 3 . Find the final volume.

Answer 1

$Work = -156.2 \;kJ = -43.34 \;Wh \\ V_1 = 0.845 \\ P = -730V + 690 \;kPa \\ W = \int VdP = V(P_1 -P_2) \\ P_1 = -730 \times 0.845 + 690 = 73.15 \;kPa \\ P_2 = -730V_2 + 690 \\ W=V(P_1-P_2) \\ -43.34 = 0.845(73.15 -P_2) \\ P_2 = 124.439 \;kPa \\ 124.439 = -730V_2 + 690 \\ V_2 = 0.775 \;m^3$