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Chapter 9: Problem 1298

The ratio of the vapor densities of two gases at a given temperature is $9: 8$, The ratio of the rms velocities of their molecule is (A) \(3: 2 \sqrt{2}\) (B) \(2 \sqrt{2}: 3\) (C) \(9: 8\) (D) \(8: 9\)

Short Answer

Expert verified
The ratio of the rms velocities of the two gases is \(\dfrac{v_{1_{rms}}}{v_{2_{rms}}} = \dfrac{2\sqrt{2}}{3}\), so the correct answer is (B) \(2 \sqrt{2}: 3\).

Step by step solution

01

Set up the ratios of vapor densities and rms velocities

We are given the ratio of vapor densities as \[\dfrac{\rho_1}{\rho_2} = \dfrac{9}{8}\] Our aim is to find the ratio of the rms velocities of the gases: \[\dfrac{v_{1_{rms}}}{v_{2_{rms}}}\]
02

Use the formulas to express the given ratio

Express the given ratio of vapor densities with the formula: \[\dfrac{\dfrac{P_1M_1}{RT_1}}{\dfrac{P_2M_2}{RT_2}} = \dfrac{9}{8}\] Since the temperatures and pressures are given to be equal for both gases, we can rewrite the equation as: \[\dfrac{M_1}{M_2} = \dfrac{9}{8}\]
03

Use the rms velocity formula to express the desired ratio

Express the desired ratio of rms velocities using the rms velocity formula: \[\dfrac{\sqrt{\dfrac{3R T_1}{M_1}}}{\sqrt{\dfrac{3R T_2}{M_2}}} = \dfrac{v_{1_{rms}}}{v_{2_{rms}}}\] Since the temperatures are equal, we can rewrite the equation as: \[\dfrac{\sqrt{\dfrac{M_2}{M_1}}}{\sqrt{\dfrac{M_1}{M_2}}} = \dfrac{v_{1_{rms}}}{v_{2_{rms}}}\]
04

Combine the equations and solve for the desired ratio

Using the ratio of molar masses from Step 2, we have: \[\dfrac{\sqrt{\dfrac{8}{9}}}{\sqrt{\dfrac{9}{8}}} = \dfrac{v_{1_{rms}}}{v_{2_{rms}}}\] Simplifying the expression gives: \[\dfrac{\sqrt{8}}{\sqrt{9}} = \dfrac{2\sqrt{2}}{3}\] So, the ratio of the rms velocities of the two gases is: \[\dfrac{v_{1_{rms}}}{v_{2_{rms}}} = \dfrac{2\sqrt{2}}{3}\] Therefore, the correct answer is (B) \(2 \sqrt{2}: 3\).

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Most popular questions from this chapter

At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are in the ratio (A) \(1: 4\) (B) \(1: 16\) (C) \(16: 1\) (D) \(4: 1\)

The temperature of an ideal gas is increased from \(27^{\circ} \mathrm{C}\) to \(927^{\circ} \mathrm{C}\). The root mean square speed of its molecules becomes (A) Four times (B) One-fourth (C) Half (D) Twice

The rms. speed of the molecules of a gas in a vessel is $400 \mathrm{~ms}^{-1}$. If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be (A) \(800 \mathrm{~ms}^{-1}\) (B) \(200 \mathrm{~ms}^{-1}\) (C) \(400 \sqrt{2} \mathrm{~ms}^{-1}\) (D) \(400 \mathrm{~ms}^{-1}\)

The rms velocity of gas molecules is \(300 \mathrm{~ms}^{-1}\). The rms velocity of molecules of gas with twice the molecular weight and half the absolute temperature is (A) \(300 \mathrm{~ms}^{-1}\) (B) \(150 \mathrm{~ms}^{-1}\) (C) \(600 \mathrm{~ms}^{-1}\) (D) \(75 \mathrm{~ms}^{-1}\)

The molecules of a given mass of a gas have a rms velocity of $200 \mathrm{~m} / \mathrm{s}\( at \)27^{\circ} \mathrm{C}\( and \)1.0 \times 10^{5} \mathrm{Nm}^{-2}\( pressure when the temperature is \)127^{\circ} \mathrm{C}$ and pressure is \(0.5 \times 10^{5} \mathrm{Nm}^{-2}\), the rms velocity in \(\mathrm{m} / \mathrm{s}\) will be (A) \(100 \sqrt{2}\) (B) \([(100 \sqrt{2}) / 3]\) (C) \([(400) / \sqrt{3}]\) (D) None of these
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