Two classrooms, labeled A and B, are of equal volume and are connected by an open door. Classroom A is warmer than classroom B (maybe it has a sunny window). Derive a formula that relates the masses of air in each room MA and MB to the temperatures TA and TB. Which room contains a greater mass of air

According to the given data, the volume of both rooms is the same, so let's consider both volumes as V. An open door connects both the room, which infer that pressure for both the rooms is also the same, let it be P.

Considering the above conditions, the general gas equation for the air in both rooms can be given as:

[tex]PV=n_{A}RT_{A}\\ \\ PV=n_{B}RT_{B}[/tex]

Here, n represents the moles of air, R is the gas constant, and T is the temperature. Taking the ratio of both the above equations we get

[tex]\frac{PV}{PV}= \frac{n_{A}RT_{A}}{n_{B}RT_{B}}\\[/tex]

moles of the gas are mass per molecular mass, as the molecular mass is the same in both the cases so n can be replaced by m, as follows

[tex]\frac{PV}{PV}= \frac{m_{A}RT_{A}}{m_{B}RT_{B}}\\[/tex]

By simplifying we get,

[tex]\frac{m_{A}}{m_{B}}=\frac{T_{B}}{T_{A}}[/tex]

According to the given conditions, TA is greater than TB. So from the derived relation, it can be seen that the mass of air in room B is greater than the mass of air in room A.