Question 5.5: Calculate the volume occupied by 0.845 mol of nitrogen gas a...

PV = nRT
V = \frac{nRT}{P}
V =\frac{0.845 \cancel{mol} \times 0.08206\frac{ L .\cancel{atm}}{ \cancel{mol} . \cancel{K}} \times 315\cancel{ K}}{ 1.37 \cancel{atm}}
= 15.9 L

CHECK The units of the answer are correct. The magnitude of the answer (15.9 L) makes sense because, as you will see in the next section, 1 mole of an ideal gas under standard temperature and pressure (273 K and 1 atm) occupies 22.4 L.
Although the temperature and pressure in this problem are not standard, they are close enough for a ballpark check of the answer. Since this gas sample contains 0.845 mol, a volume of 15.9 L is reasonable.