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?

Jan 23, 2017

${V}_{2} \cong 5 \cdot L$
$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, from the $\text{combined gas law}$.
And thus ${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2}$, 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