# What is the volume of 5.6 moles of nitrogen gas at STP?

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 = 5.60 moles
R = 0.0821 $\frac{a t m L}{m o l K}$
T = 273 K

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

$V = \left(\frac{\left(5.60 m o l\right) \left(0.0821 \frac{a t m L}{m o l K}\right) \left(273 K\right)}{1 a t m}\right)$

#V = 125.51 L

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

$5.60 m o l x \frac{22.4 L}{1 m o l} = 125.44 L$

In both cases the significant digit count is three figures which means the acceptable solution is $125 L$.