# What is the pressure in atmospheres of 0.246 gram of hydrogen gas occupying a volume of 0.0500 liters at 21.0°C?

Oct 10, 2016

The pressure of $\text{H"_2}$ gas will be 58.9 atm, rounded to three significant figures.

#### Explanation:

You need the ideal gas law to answer this question. First convert the mass of hydrogen gas $\left(\text{H"_2}\right)$ to moles by dividing the given mass by its molar mass.

Molar mass $\text{H"_2}$$=$$\left(1.008 \text{g/mol"xx2=2.016"g/mol}\right)$.

Moles $\text{H"_2}$

$\left(0.246 \cancel{\text{g H"_2)/(2.016cancel"g/mol H"_2}}\right)$$=$$\text{0.122 mol H"_2}$

Now you need to determine your known and unknown variables.

Known
$V = \text{0.0500 L}$
$n = \text{0.122 mol}$
$R = \text{0.082057338 L atm K"^(-1) "mol"^(-1)}$
$T = \text{21.0"^@"C"+"273.15=294.2K}$

Unknown
$P$

Solution
Rearrange the ideal gas equation to isolate $P$ and solve.

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

P=((0.122cancel"mol")xx(0.082057338 cancel"L" "atm" cancel"K"^(-1) cancel"mol"^(-1))xx(294.2cancel"K"))/(0.0500cancel"L")="58.9 atm"
rounded to three significant figures