Answer
The new pressure is $3.3\times 10^5~Pa$
Work Step by Step
We can find an expression for the original pressure:
$P_1~V_1 = nRT_1$
$P_1 = \frac{nRT_1}{V_1}$
We can find an expression for the final pressure:
$P_2~V_2 = nRT_2$
$P_2 = \frac{nRT_2}{V_2}$
We can divide $P_2$ by $P_1$:
$\frac{P_2}{P_1} = \frac{\frac{nRT_2}{V_2}}{\frac{nRT_1}{V_1}}$
$P_2 = \frac{V_1~T_2~P_1}{V_2~T_1}$
$P_2 = \frac{(1.2~m^3)(500~K)(1.0\times 10^5~Pa)}{(0.60~m^3)(300~K)}$
$P_2 = 3.3\times 10^5~Pa$
The new pressure is $3.3\times 10^5~Pa$
