Kinetic Theory of Gases: Practice Problem & Solution
Match Column - I and Column - II and choose the correct match from the given choices. Column-I (A) Root mean square speed of gas molecules (B) Pressure exerted by ideal gas (C) Average kinetic energy of a molecule (D) Total internal energy of 1 mole of a diatomic gas Column-II (P) \(\frac{1}{3} n m \bar{v}^2\) (Q) \(\sqrt{\frac{3RT}{M}}\) (R) \(\frac{5}{2} RT\) (S) \(\frac{3}{2} k_B T\)
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:
By kinetic theory: \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\) -> (A)-(Q); Pressure \(P = \frac{1}{3} nm\bar{v}^2\) -> (B)-(P); Average KE \(= \frac{3}{2}k_BT\) -> (C)-(S); and Internal energy for diatomic gas \(= \frac{5}{2}RT\) -> (D)-(R).
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