# A steel tank with a volume of 9.583 L contains N_2 gas under a pressure of 4.972 atm at 31.8°C. How many moles of N_2 are in the tank?

$n \cong 2 \cdot m o l$
For the old Ideal Gas equation, $n = \frac{P V}{R T}$...
And so $n = \frac{4.972 \cdot a t m \times 9.583 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \cdot 305 \cdot K}$
$= 1.90 \cdot m o l$