Essential University Physics: Volume 1 (3rd Edition)

(a) We know that $PV=nRT$ This can be rearranged as: $V=\frac{nRT}{P}$ We plug in the known values to obtain: $V=\frac{(2.0)(8.314)(250)}{(1.5)(1.01\times 10^5)}$ $V=0.027m^3$ (b) As $PV=nRT$ This can be rearranged as: $T=\frac{PV}{nR}$ We plug in the known values to obtain: $T=\frac{(4.0)(1.01\times 10^5)(\frac{0.027}{2})}{(2.0)(8.314)}$ $T=330K$