# How many moles of gas at 5.60 atm and 300 K will occupy a volume of 10.0 L?

Dec 19, 2015

$2.28$ $\text{mol}$

#### Explanation:

Use the ideal gas law equation

$P V = n R T$

where

$P = \text{pressure} = 5.60$ $\text{atm}$
$V = \text{volume} = 10.0$ $\text{L}$
$n = \text{moles}$
$R = \text{ideal gas constant} = 0.082$ $\left(\text{L atm")/("K mol}\right)$
$T = \text{temperature} = 300$ $\text{K}$

Since we want to find the amount of moles, we can rearrange the equation by dividing by $R T$.

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

$n = \frac{5.60 \times 10.0}{.082 \times 300} = 2.28$ $\text{mol}$