Kinetic energy column matching – Rankers Physics

Kinetic Theory of Gases: Practice Problem & Solution

Consider a sample of \(n\) moles of rigid diatomic gas. Match the columns and tick the correct option (symbols have their usual meanings): 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)-(P); (D)-(S)
(A)-(Q); (B)-(R); (C)-(S); (D)-(P)
(A)-(Q); (B)-(R); (C)-(P); (D)-(S)
(A)-(R); (B)-(Q); (C)-(S); (D)-(P)

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 of \(n\) moles is \(\frac{3}{2}nRT\) (R). Rotational KE is \(nRT\) (Q). KE per mole of diatomic gas is \(\frac{5}{2}RT\) (S). KE per molecule is \(\frac{5}{2}k_B T\) (P). Thus, (A)-(R), (B)-(Q), (C)-(S), (D)-(P).

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): In an ideal monoatomic gas, The Internal energy of gas is equal to translational Kinetic energy of all its molecules Reason (R): The Internal energy may get contributes from Translational, Rotatory, vibrationally as well as from the Potential energy corresponding to the molecular force.  
2. Assertion (A): There is no change in internal energy for ideal gas at constant temperature. Reason (R): Internal energy of an ideal gas is a function of temperature only.  
3. Given below are two statements: Assertion (A): The average translational kinetic energy per molecule of gas for various gases at the same temperature is the same. Reason (R): At a given temperature, all molecules of a gas move with nearly the same speed.  
4. Assertion (A): Molar heat capacity of an ideal monoatomic gas at constant volume is a constant at all temperatures. Reason (R): As the temperature of an monoatomic ideal gas is increased, number of degrees of freedom of gas molecules remains constant.  
5. Assertion (A): For gas molecules absolute zero temperature is not the temperature of zero energy. Reason (R): Only the kinetic energy of the molecules is represented by temperature.