# What is the volume occupied by 0.355 mole of nitrogen gas at STP?

##### 1 Answer
Jun 10, 2014

There are two ways to calculate this if we assume STP of 1 atm and 273 K for the pressure and temperature.

We can use the Ideal Gas Law equations PV = nRT
P = 1 atm
V = ???
n = 0.355 moles
R = 0.0821 $\frac{a t m L}{m o l K}$
T = 273 K

$P V = n R T$ can be $V = \frac{n R T}{P}$

$V = \left(\frac{\left(0.355 m o l\right) \left(0.0821 \frac{a t m L}{m o l K}\right) \left(273 K\right)}{1 a t m}\right)$

V = 7.96 L

The second method is to us Avogadro's Volume at STP $22.4 L = 1 m o l$

 0.355 mol × (22.4 L)/(1 mol) = 7.95 L#

I hope this was helpful.
SMARTERTEACHER