The internal energy of air in a room of dimensions 4 m × 4 m × 3 m at atmospheric pressure 1×10⁵ Pa is N×10⁶ J. Treat air as an ideal diatomic gas. Find N.
Physics · Thermodyanmics · kinetic therory of gaseous
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Question

The internal energy of air in a room of dimensions 4 m × 4 m × 3 m at atmospheric pressure 1×10⁵ Pa is N×10⁶ J. Treat air as an ideal diatomic gas. Find N.

G1 Physics · Accepted Answer

For an ideal diatomic gas at ordinary temperature, the number of active degrees of freedom is f = 5. Internal energy is U = (f/2)nRT. Using nRT = PV, U = (f/2)PV = (5/2)PV. Room volume: V = 4×4×3 = 48 m³. Therefore, U = (5/2)(1×10⁵)(48) = 2.5×48×10⁵ = 120×10⁵ = 12×10⁶ J. Hence N = 12.