# What is the volume occupied by 10.8 g of argon gas at a pressure of 1.30 atm and a temperature of 408 K?

Nov 23, 2015

The volume of argon will be 6.96 L.

#### Explanation:

Use the ideal gas law $P V = n R T$, where $n$ is moles and $R$ is the gas constant.

Since we have mass and need moles, we must divide the given mass of argon by its molar mass (atomic weight on the periodic table in g/mol). The molar mass of argon is 39.948 g/mol.

$\text{Ar} :$$10.8 \text{g Ar"xx(1"mol Ar")/(39.948"g Ar")="0.27035 mol Ar}$

I am keeping a couple of guard digits to reduce rounding errors. I will round the final answer to three significant figures.

Ideal Gas Law

Given/Known
$P = \text{1.30 atm}$
$n = \text{0.27035 mol}$
$R = \text{0.082057 L atm K"^(-1) "mol"^(-1)}$
https://en.wikipedia.org/wiki/Gas_constant
$T = \text{408 K}$

Unknown
$V$

Equation
$P V = n R T$

Solution
Rearrange the equation to isolate volume, $V$, and solve.

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

V=(0.27035cancel"mol"xx0.082057"L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1))xx408cancel"K")/(1.30cancel"atm")="6.96 L Ar"