The internal energy of air in 4 m × 4 m × 3 m sized room at 1 atmospheric pressure will be___________________× 10⁶ J
(Consider air as diatomic molecule)

Solution

Physics sir · 4 steps

No temperature given, no problem: for an ideal gas nRT is just PV, and the room gives you both.

1. Count degrees of freedom. A diatomic molecule at room temperature has 3 translational and 2 rotational degrees of freedom, so f = 5. Equipartition gives each degree of freedom 1/2k_BT per molecule.
2. Replace nRT with PV. For n moles, U = f/2nRT. The ideal gas equation says nRT = PV, so the unknown temperature and moles drop out.

   U = 5/2 PV

3. Put in numbers. V = 4 × 4 × 3 = 48 m³ and P = 1.013 × 10⁵ Pa. So U = 2.5 × 1.013 × 10⁵ × 48 = 1.216 × 10⁷ J.
4. So. U ≈ 12.2 × 10⁶ J, which gives 12.

The common slip: 7. Using f = 3 as for a monatomic gas gives 3/2PV ≈ 7.3 × 10⁶ J, so 7, not 12. Air is diatomic, so the two rotational degrees of freedom count too.

The idea to keep. Internal energy of an ideal gas is f/2PV. When a question gives pressure and volume but no temperature, do not hunt for T; just multiply PV by f/2.

Answer from the NTA final answer key. Solution by PrepWiser.