Question

Please help!!!!!

A 6.10m × 7.60m × 2.90m room contains air at 20∘C. what is the rooms thermal energy?

Answer 1

You can solve for average thermal energy of the room by using this formula KEav=(3/2)kT where k=1.38*10-23 and T= temperature in kelvin in your case 20+273=293k
however i am not sure how the volume of the room is related to the changing energy?

Answer 2

That's for a single molecule, I believe - a single atom, really, so if it's air you might want to add an extra degree of freedom, making it
\[\bar{K}=\frac{5}{2}kT\]
so if you assume it's an ideal gas, you can solve for the number of molecules based on
\[PV=NkT\]
\[N=\frac{PV}{kT}\]
where
\[P=1atm = 101325N/m^2\]
\[V = \text{volume of room as m}^3\]
Where N is he number of gas particles

So the total average thermal energy in the room would be the average kinetic energy multiplied by the number of molecules - then you get an answer reasonable in terms of the order of magnitude.
\[\sum K = N K_E = \frac{5}{2}PV\]

Answer 3

last term should be
\[\sum_{i=1}^N \bar{K_i} = N \bar{K} = \frac{5}{2}PV\]
where
\[\bar{K}_i \overset{i≠j}{=} \bar{K}_j\]
Sorry for sloppy conventions :P