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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 \(100 \mathrm{~K}\) and \(0.1\) atmospheric pressure, the volume helium gas is 10 liters. If volume and pressure are doubled, its temperature will change to (A) \(127 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(25 \mathrm{~K}\) (D) \(200 \mathrm{~K}\)

At a given volume and temperature the pressure of gas (A) Varies inversely as the square of its mass (B) Varies inversely as its mass (C) is independent of its mass (D) Varies linearly as its mass

The pressure and temperature of an ideal gas in a closed vessel are $720 \mathrm{kPa}\( and \)40^{\circ} \mathrm{C}\( respectively. If \)(1 / 4)^{\text {th }}$ of the gas is released from the vessel and the temperature of the remaining gas is raised to \(353^{\circ} \mathrm{C}\), final pressure of the gas is (A) \(1440 \mathrm{kPa}\) (B) \(540 \mathrm{kPa}\) (C) \(1080 \mathrm{kPa}\) (D) \(720 \mathrm{kPa}\)

The average kinetic energy per molecule of a gas at \(-23^{\circ} \mathrm{C}\) and \(75 \mathrm{~cm}\) pressure is \(5 \times 10^{-14}\) erg for \(\mathrm{H}_{2}\). The mean kinetic energy per molecule of the \(\mathrm{O}_{2}\) at \(227^{\circ} \mathrm{C}\) and \(150 \mathrm{~cm}\) pressure will be (A) \(80 \times 10^{-14} \mathrm{erg}\) (B) \(10 \times 10^{-14}\) erg (C) \(20 \times 10^{-14}\) erg (D) \(40 \times 10^{-14}\) erg

A sample of gas is at \(0^{\circ} \mathrm{C}\). To what temperature it must be raised in order to double the rms speed of molecule. (A) \(270^{\circ} \mathrm{C}\) (B) \(819^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(1090^{\circ} \mathrm{C}\)
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