Answer
(a) 69.5 K
(b) 32.4 atm
Work Step by Step
Volume $V=2.50\,L$
Amount of gas $n=\frac{175\,g}{39.948\,g/mol} =4.38\,mol$
$R=0.0821\,L\,atm\,mol^{-1}K^{-1}$
$PV=nRT$ (ideal gas law)
(a) When $P=10.0\,atm$,
$T=\frac{PV}{nR}=\frac{10.0\,atm\times2.50\,L}{4.38\,mol\times0.0821\,L\,atm\,mol^{-1}K^{-1}}$
$=69.5\,K$
(b) When $T=225\,K$,
$P=\frac{nRT}{V}$
$=\frac{4.38\times0.0821\times225}{2.50}\,atm$
$=32.4\,atm$
