To determine the internal energy of the air in a room that measures $4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m}$ at 1 atmospheric pressure, we can follow these steps using the properties of a diatomic gas:

The volume of the room, $ V $, is calculated as:

$ V = 4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m} = 48 \, \mathrm{m}^3 $

For a diatomic gas like air, the internal energy $ U $ is given by the formula:

$ U = nC_v T = n \frac{5RT}{2} $

Where:

$ n $ is the number of moles,

$ C_v $ is the molar heat capacity at constant volume,

$ R $ is the ideal gas constant, and

$ T $ is the temperature in Kelvin.

Using the ideal gas law, $ PV = nRT $, we can substitute for $ nRT $:

$ nRT = PV $

Therefore, the internal energy $ U $ becomes:

$ U = \frac{5}{2} PV $

Substituting the given values for atmospheric pressure $ P = 10^5 \, \mathrm{Pa} $ (1 atm) and volume $ V = 48 \, \mathrm{m}^3 $:

$ U = \frac{5}{2} \times 10^5 \, \mathrm{Pa} \times 48 \, \mathrm{m}^3 = 12 \times 10^6 \, \mathrm{J} $

Thus, the internal energy of the air in the room is $ 12 \times 10^6 \, \mathrm{J} $.