# If 64 grams of O_2 gas occupy a volume of 56 liters at a temperature or 25°C, what is the pressure of the gas, in atmosphere?

Nov 27, 2016

The pressure of oxygen gas is 0.87 atm.

#### Explanation:

Use the equation for the ideal gas law.

Since gas laws require temperature to be in Kelvins, the Celsius temperature will have to be converted by adding $273.15$ to the Celsius temperature.

You don't have $n$, but it can be calculated by dividing the given mass of $\text{O"_2}$ by its molar mass: $2 \times 15.999 \text{g/mol"="31.998 g/mol}$

${n}_{\text{oxygen"=(64cancel"g")/(31.998cancel"g"/"1 mol")="2.0 mol O"_2}}$

Known
$n = \text{2.0 mol O"_2}$
$V = \text{56 L}$
$T = \text{25"^@"C"+273.15="298 K}$
$R = 0.08206 \left(\text{L·atm")/("K·mol}\right)$

Unknown
$P$

Solution
Rearrange the equation to isolate $P$. Substitute the known values into the equation and solve.

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

$P = 2.0 \cancel{\text{mol"xx0.08206 (cancel"L"·"atm")/(cancel"K"·cancel"mol")xx(298cancel"K")/(56cancel"L")="0.87 atm}}$ rounded to two figures