# Why would knowing the molar volume of a gas be important?

Dec 30, 2014

The molar volume of a gas expresses the volume occupied by 1 mole of that respective gas under certain temperature and pressure conditions.

The most common example is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.

So, if you are given these values for temperature and pressure, the volume occupied by any number of moles of an ideal gas can be easily derived from knowing that 1 mole occupies 22.4 L.

$V = n \cdot {V}_{m o l a r}$

For 2 moles of a gas at STP the volume will be

$2$ $\text{moles} \cdot 22.4$ $\text{L/mol} = 44.8$ $\text{L}$

For 0.5 moles the volume will be

$0.5$ $\text{moles} \cdot 22.4$ $\text{L/mol} = 11.2$ $\text{L}$, and so on.

The molar volume of a gas is derived from the ideal gas law $P V = n R T$:

$P V = n R T \to V = \frac{n R T}{P} \to \frac{V}{n} = \frac{R T}{P}$

Let's say you were given a temperature of 355 K and a pressure of 2.5 atm, and asked to determine the gas' molar volume at these conditions. Since molar volume refers to the volume occupied by 1 mole, you'd get

$\frac{V}{\text{1 mole}} = \frac{0.082 \frac{L \cdot a t m}{m o l \cdot K} \cdot 355 K}{2.5 a t m} = 11.6$ $\text{L/mol}$

This is how much volume 1 mole occupies at 355 K and 2.5 atm. It becomes clear that the volume occupied by any number of moles at these conditions can be easily determined:

$2$ $\text{moles} \cdot 11.6$ $\text{L/mol} = 23.2$ $\text{L}$

$0.5$ $\text{moles} \cdot 11.6$ $\text{L/mol} = 5.8$ $\text{L}$, and so on.

As a conclusion, knowing a gas' molar volume at a certain temperature and a certain pressure can simplify the calculation of the volume occupied by any number of moles of that respective gas.

Good explanation, good figures here: http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_Lower%27s_Chem1/06._Properties_of_Gases/6.3%3A_Dalton%27s_Law