## College Physics (4th Edition)

The new pressure is $3.3\times 10^5~Pa$
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 $3.3\times 10^5~Pa$