What is the density of #O_2# at 298 K and 0.987 atm?

1 Answer
Dec 29, 2016

Answer:

#rho_(O_2)~=1*g*L^-1#

Explanation:

We start with the ideal gas equation.....

#PV=nRT; P/(RT)=n/V=("mass"/"molar mass")/V#.

And thus #P/(RT)xx"molar mass"="mass"/V=rho#

So we simply fill in the blanks,

#rho=(1*cancel(atm)xx32.00*g*cancel(mol^-1))/(0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx298*cancel(K)#

#rho=??*g*L^-1#

We get the required units of density.