Match the energy columns for a diatomic gas – Rankers Physics

Kinetic Theory of Gases: Practice Problem & Solution

Column I Column II (A) Total translational kinetic energy (P) $\frac{5}{2} K_B T$ (B) Total rotational kinetic energy (Q) $nRT$ (C) Total kinetic energy per mole (R) $\frac{3}{2} nRT$ (D) Total kinetic energy per molecule (S) $\frac{5}{2} RT$
(A)-(R); (B)-(Q); (C)-(S); (D)-(P)
(A)-(Q); (B)-(R); (C)-(S); (D)-(P)
(A)-(Q); (B)-(R); (C)-(P); (D)-(S)
(A)-(R); (B)-(Q); (C)-(P); (D)-(S)

Solution Explained:

To solve this problem, we apply the core principles of Kinetic Theory of Gases. Understanding the underlying formula is key to arriving at the correct answer below:

Translational KE is \(\frac{3}{2}nRT\). Rotational KE of a rigid diatomic gas is \(nRT\). Total KE per mole is \(\frac{5}{2}RT\) and total KE per molecule is \(\frac{5}{2}K_B T\).

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