The internal energy of air in $4 \mathrm{~m} \times 4 \mathrm{~m} \times 3 \mathrm{~m}$ sized room at 1 atmospheric pressure will be___________________$\times 10^6 \mathrm{~J}$

(Consider air as diatomic molecule)

<p>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:</p> <p>The volume of the room, $ V $, is calculated as:</p> <p>$ V = 4 \, \mathrm{m} \times 4 \, \mathrm{m} \times 3 \, \mathrm{m} = 48 \, \mathrm{m}^3 $</p> <p>For a diatomic gas like air, the internal energy $ U $ is given by the formula:</p> <p>$ U = nC_v T = n \frac{5RT}{2} $</p> <p>Where:</p> <p><p>$ n $ is the number of moles,</p></p> <p><p>$ C_v $ is the molar heat capacity at constant volume,</p></p> <p><p>$ R $ is the ideal gas constant, and</p></p> <p><p>$ T $ is the temperature in Kelvin.</p></p> <p>Using the ideal gas law, $ PV = nRT $, we can substitute for $ nRT $:</p> <p>$ nRT = PV $</p> <p>Therefore, the internal energy $ U $ becomes:</p> <p>$ U = \frac{5}{2} PV $</p> <p>Substituting the given values for atmospheric pressure $ P = 10^5 \, \mathrm{Pa} $ (1 atm) and volume $ V = 48 \, \mathrm{m}^3 $:</p> <p>$ U = \frac{5}{2} \times 10^5 \, \mathrm{Pa} \times 48 \, \mathrm{m}^3 = 12 \times 10^6 \, \mathrm{J} $</p> <p>Thus, the internal energy of the air in the room is $ 12 \times 10^6 \, \mathrm{J} $.</p>