Answer
59.41 kg
Work Step by Step
Let's take,
Mass of the air in the room = $m$
Mass of the nitrogen = $m_{n}$
Mass of the oxygen = $m_{o}$
Number of moles of nitrogen = $n_{n}$
Number of moles of oxygen = $n_{o}$
Molecular mass of nitrogen = $M_{n}$
Molecular mass of oxygen = $M_{o}$
We can write,
$m=m_{n}+m_{o}=n_{n}M_{n}+n_{o}M_{o}$
We can express the above equation by using $f$ to denote the fraction of a species that is present and $n_{total}$ to denote the total number of moles,
$m=f_{n}n_{total}M_{n}+f_{o}n_{total}M_{o}-(1)$
Let's apply the ideal gas law $PV=nRT$ to the system.
$PV=nRT=>n_{total}=\frac{PV}{RT}-(2)$
(2)=>(1),
$m=(f_{n}M_{n}+f_{o}M_{o})n_{total}=(f_{n}M_{n}+f_{o}M_{o})(\frac{PV}{RT})$
Let's plug known values into this equation.
$m=\{0.79(28\space g/mol)+0.21(32\space g/mol)\}(\frac{(1.01\times10^{5}Pa)(2.5\times4\times5\space m^{3})}{(8.13\space J/mol\space K)(295\space K)})$
$m=59410.6\space g=59.41\space kg$
mass of the air = 59.41 kg