A $6.0\mathrm{m}\times 8.0\mathrm{m}\times 3.0\mathrm{m}$ room contains air at ${20}^{O}C$. What is the room's thermal energy?

In the work domain, the potential energy between the bonds is zero, and the kinetic energy of a monatomic gas is translational kinetic energy.. As a result, a diatomic gas's thermal energy is $n$moles is given by equation $(20.37)$in the form

$E_{\mathrm{th}}=\frac{5}{2}nRT$

The link between the four state variables for an ideal gas, which are container volume, pressure,$P$ and temperature, is depicted by the ideal gas law, and that the gas is exerted, temperature $T$of the gas and the number of moles $n$of the gas in the container. Ideal gas law is given by equation (18.13) in the form

$pV=nRT$

We substitute the term $nRT$with $PV$into equation $\left(1\right)$, so the thermal energy would be

$E_{\mathrm{th}}=\frac{5}{2}pV$

The pressure in the room is at the atmospheric pressure $p=1.013\times {10}^5\mathrm{Pa}$. The volume of the air in the room is calculated by

$V=6.0\mathrm{m}\times 8.0\mathrm{m}\times 3.0\mathrm{m}=144{\mathrm{m}}^3$

Now, we plug the values for $p$and $v$into equation (3) to get

$E_{\mathrm{th}}=\frac{5}{2}pV$

$=\frac{5}{2}\left(1.013\times {10}^5\mathrm{Pa}\right)\left(144{\mathrm{m}}^3\right)$

$=3.6\times {10}^7\mathrm{J}$