Answer
131 $\mathrm{g}$ of xenon gas could not exert a pressure of 20 atm. It would exert a pressure of 21.65 $\mathrm{atm}$
Work Step by Step
In this exercise we have to answer: could 131 $\mathrm{g}$ of xenon gas in a vessel of
volume $\mathrm{V}=1.0 \mathrm{L}$ exert a pressure of 20 atm at the temperature of $\mathrm{T}=298$
$\mathrm{K}$ if it behaved as a perfect gas:
$\begin{aligned} \mathrm{n} &=\frac{131 \mathrm{gm}}{131.0293 \mathrm{gm} \mathrm{mol}} \\ &=0.997 \mathrm{mol} \end{aligned}$
Now calculate the pressure by using the expression:
$P=\frac{n R T}{V}$
Values:
$\mathbf{n}=0.997$ mol
$\mathbf{R}=0.0821 \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1}-$ gas constant
$\mathrm{T}=298 \mathrm{K}$ - temperature
$\mathrm{V}=1.0 \mathrm{L}$ - volume
$\begin{aligned} P &=\frac{n R T}{V} \\ &=\frac{0.997 \mathrm{mol} \cdot 0.0821 \mathrm{Latm} \mathrm{mol}^{-1} \cdot 298 \mathrm{K}}{1.0 \mathrm{L}} \\ &=24.4 \mathrm{atm} \end{aligned}$
This value is higher than 20.0 atm so it isn't a perfect gas
Now we will calculate the pressure when xenon gas behaves as a vander waals gas
$$
\begin{aligned} P &=\frac{R T}{V_{m}-b}-\frac{a}{V_{W}^{2}} \\ &=\frac{\left(0.0821 \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\right)(298 \mathrm{K})}{(1.0 \mathrm{L} / \mathrm{mol})^{2}} \\ &=(25.796-4.137) \mathrm{atm} \\ &=21.65 \mathrm{atm} \end{aligned}
$$
