Question

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?

Answer 1

The pressure of oxygen gas is 0.87 atm.

Explanation:

Use the equation for the ideal gas law.
http://imgarcade.com/1/ideal-gas-law-equation/
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 `"O"_2"` by its molar mass: `2xx15.999"g/mol"="31.998 g/mol"`

`n_"oxygen"=(64cancel"g")/(31.998cancel"g"/"1 mol")="2.0 mol O"_2"`

Known
`n="2.0 mol O"_2"`
`V="56 L"`
`T="25"^@"C"+273.15="298 K"`
`R=0.08206 ("L·atm")/("K·mol")`

Unknown
`P`

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

`P=(nRT)/V`

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