# What is the volume occupied by one mole of helium at 0 C and 1 atm pressure?

Feb 23, 2017

$\text{22.4 L}$

#### Explanation:

The conditions for temperature and pressure provided to you actually correspond to the old definition of STP (Standard Pressure and Temperature).

Under these specific conditions, $1$ mole of any ideal gas occupies $\text{22.4 L}$. This value is known as the molar volume of a gas.

You can show that this is the case by using the ideal gas law equation, which looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange the ideal gas law equation to

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

Plug in your values to find -- do not forget to convert the temperature from degrees Celsius to Kelvin

V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))

$\frac{V}{n} = {\text{22.4 L mol}}^{- 1}$

This means that under these conditions for pressure and temperature, you get $\text{22.4 L}$ for every mole of an ideal gas present in a sample.

SIDE NOTE STP conditions are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these specific conditions, the molar volume of a gas is equal to ${\text{22.7 L mol}}^{- 1}$.