A 2.94*L volume of gas at 294*K and under 69.6*kPa pressure enclosed in a piston, is cooled to 257*K and the pressure reduced to 35.6*kPa. What is the new volume?

1 Answer
Jan 23, 2017

V_2~=5*L

Explanation:

(P_1V_1)/T_1=(P_2V_2)/T_2, from the "combined gas law".

And thus V_2=(P_1V_1)/T_1xxT_2/P_2, which expression CLEARLY has the units of volume.

So V_2=(69.9*cancel(kPa)xx2.94*L)/(294*cancel(K))xx(257*cancel(K))/(35.6*cancel(kPa))=??L