Question 572: Thermodynamics & KTG

Question

The internal energy of air in $4 \mathrm{~m} 	imes 4 \mathrm{~m} 	imes 3 \mathrm{~m}$ sized room at 1 atmospheric pressure will be___________________$	imes 10^6 \mathrm{~J}$
(Consider air as diatomic molecule)

Accepted Answer

To determine the internal energy of the air in a room that measures $4 \, \mathrm{m} 	imes 4 \, \mathrm{m} 	imes 3 \, \mathrm{m}$ 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} 	imes 4 \, \mathrm{m} 	imes 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} 	imes 10^5 \, \mathrm{Pa} 	imes 48 \, \mathrm{m}^3 = 12 	imes 10^6 \, \mathrm{J} $
Thus, the internal energy of the air in the room is $ 12 	imes 10^6 \, \mathrm{J} $.

Thermodynamics & KTG

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)

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