# If 28.0 g of methane gas (CH_4) are introduced into an evacuated 2.00-L gas cylinder at a temperature of 35°C, what is the pressure inside the cylinder?

Feb 22, 2016

The pressure of the methane gas will be 22.2 atm.

#### Explanation:

Use the ideal gas law with the formula $P V = n R T$, where $n$ is moles, $R$ is the gas constant, and $T$ is temperature in Kelvins.

Determine the mole of $\text{CH"_4}$ by dividing the given mass by its molar mass, $\text{16.04246 g/mol}$ https://pubchem.ncbi.nlm.nih.gov/compound/297

$28.0 \cancel{\text{g CH"_4xx(1"mol CH"_4)/(16.04246cancel"g CH"_4)="1.7454 mol CH"_4}}$

Ideal Gas Law

Given/Known
$V = \text{2.00 L}$
$n = \text{1.7454 mol}$
$R = {\text{0.082057338 L atm K"^(-1) "mol}}^{- 1}$
$T = \text{35"^@"C"+273.15="308 K}$

Unknown
Pressure, $P$

Solution
Rearrange the formula to isolate $P$. Substitute the given values into the formula and solve.

$P V = n R T$

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

P=((1.7454"mol") xx (0.082057338"L atm K"^(-1) "mol"^(-1)) xx (308"K"))/(2.00"L")="22.2 atm" rounded to three significant figures