#### Answer

$p_{2}$ = 2.6 atm

#### Work Step by Step

We know that a rigid container holds Hydrogen gas. Therefore
$V_{2}$ = $V_{1}$ .
We can use the ideal gas law
$\frac{p_{1}V_{1}}{T_{1}}$ =$\frac{p_{2}V_{2}}{T_{2}}$,
where
$p_{1}$ is the initial pressure,
$p_{2}$ is the final pressure,
$T_{1}$ is the initial temperature,
$T_{2}$ is the final temperature,
$V_{1}$ is the initial volume, and
$V_{2}$ is the final volume.
To find the final pressure( $p_{2}$), we are given :
$p_{1}$ - 3.0 atm
$T_{1}$ - 20$^{\circ}$C which must be converted to Kelvins(K) = 293K
$T_{2}$ = -20$^{\circ}$C = 253K
We convert to Kelvins by adding 273 to the Temperature in Celsius:
$T_{K}$ = 20$^{\circ}$C + 273 = 293K.
Now we are ready to solve for $p_{2}$
So $p_{2} = \frac{p_{1}V_{1}T_{2}}{V_{2}T_{1}}$ = $\frac{p_{1}V_{1}T_{2}}{V_{1}T_{1}}$ = (3 atm) $\frac{253}{293}$ = 2.6 atm