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?

1 Answer
Nov 27, 2016

Answer:

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