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Chapter 13: Problem 100

Show that, in two gases at the same temperature, the rms speeds are inversely proportional to the square root of the molecular masses: $$ \frac{\left(v_{\mathrm{rms}}\right)_{1}}{\left(v_{\mathrm{rms}}\right)_{2}}=\sqrt{\frac{m_{2}}{m_{1}}} $$

Short Answer

Expert verified
Answer: The rms speeds of two gases at the same temperature are inversely proportional to the square root of their molecular masses, given by the equation: $$\frac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2} = \sqrt{\frac{m_2}{m_1}}$$

Step by step solution

01

Recall the equation for the rms speed.

The root-mean-square (rms) speed of gas particles is given by the following equation: $$ v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} $$ where \(v_{\mathrm{rms}}\) is the rms speed, \(k\) is the Boltzmann's constant, \(T\) is the temperature, and \(m\) is the molecular mass of the gas particles.
02

Write down the equation for the two gases.

Let's consider two gases with rms speeds \((v_{\mathrm{rms}})_1\) and \((v_{\mathrm{rms}})_2\), molecular masses \(m_1\) and \(m_2\), and the same temperature \(T\). We can write down their respective rms speed equations: $$ \begin{aligned} (v_{\mathrm{rms}})_1 = \sqrt{\frac{3kT}{m_1}} \\ (v_{\mathrm{rms}})_2 = \sqrt{\frac{3kT}{m_2}} \end{aligned} $$
03

Set up a ratio of the rms speeds.

Divide the equation for \((v_{\mathrm{rms}})_1\) by the equation for \((v_{\mathrm{rms}})_2\): $$ \frac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2} = \frac{\sqrt{\frac{3kT}{m_1}}}{\sqrt{\frac{3kT}{m_2}}} $$
04

Simplify the ratio.

Notice that \(3kT\) is common in both the numerator and the denominator, so it cancels out. The ratio then simplifies to: $$ \frac{(v_{\mathrm{rms}})_1}{(v_{\mathrm{rms}})_2} = \sqrt{\frac{m_2}{m_1}} $$ Therefore, the rms speeds of two gases at the same temperature are inversely proportional to the square root of their molecular masses, as given by the equation.

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