Kinetic Theory of Gases: Practice Problem & Solution
The molecules of a given mass of a gas have r.m.s velocity of $200\text{ ms}^{-1}$ at $27^{\circ}\text{C}$ and $1.0 \times 10^5\text{ Nm}^{-2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ}\text{C}$ and $0.05 \times 10^5\text{ Nm}^{-2}$, the r.m.s. velocity of its molecules in $\text{ms}^{-1}$ is: (2016 - I)
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:
RMS velocity depends only on temperature: $v_{rms} \propto \sqrt{T}$. Let $v_1$ and $v_2$ be rms speeds at $T_1 = 27^{\circ}\text{C} = 300\text{ K}$ and $T_2 = 127^{\circ}\text{C} = 400\text{ K}$. $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}}$. So $v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}\text{ ms}^{-1}$.
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