Answer
The pressure will be 2.6 atm
Work Step by Step
The initial temperature is 293 K and the final temperature is 253 K. We can express $T_2$ in terms of $T_1$.
$\frac{T_2}{T_1} = \frac{253~K}{293~K}$
$T_2 = \frac{253~K}{293~K}~T_1$
$T_2 = 0.8635~T_1$
We can find an expression for the initial pressure which is 3.0 atm.
$P_1V = nRT_1$
$P_1 = \frac{nRT_1}{V}$
We can find the final pressure.
$P_2V = nRT_2$
$P_2 = \frac{nRT_2}{V}$
$P_2 = \frac{nR(0.8635~T_1)}{V}$
$P_2 = (0.8635)~\frac{nRT_1}{V}$
$P_2 = (0.8635)~P_1$
$P_2 = (0.8635)(3.0~atm)$
$P_2 = 2.6~atm$
The pressure will be 2.6 atm.
