# What is the molar volume of 46.00 grams of nitrogen gas?

Dec 28, 2014

Molar volume problems should provide some information about temperature and pressure, that is the only way to determine a substance's molar volume.

Starting from the ideal gas law, $P V = n R T$, the expression of an ideal gas' molar volume is

$P V = n R T \to \frac{n R T}{P} \to \frac{V}{n} = \frac{R T}{P} = {V}_{m o l a r}$

If we take STP conditions - 273.15 K and 1.00 atm, the volume occupied by 1 mole of gas will be

$\frac{V}{1 m o l e} = \frac{0.082 \frac{L \cdot a t m}{m o l \cdot K} \cdot 273.15 K}{1.00 a t m} = 22.4 \frac{L}{m o l} = {V}_{m o l a r}$

So, if we have more moles of a gas, the volume occupied will be bigger at STP:

${n}_{{N}_{2}} = \frac{m}{m o l a r . m a s s} = \frac{46.0 g}{14.0 \frac{g}{m o l}} = 3.29$ moles. So,

$V = n \cdot {V}_{m o l a r} = 3.29 m o l e s \cdot 22.4 \frac{L}{1 m o l} = 73.7 L$