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Chapter 9: Problem 1305

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) \(4: 1\)

Short Answer

Expert verified
The root mean square velocities of Oxygen and Hydrogen molecules are in the ratio \(1:4\). The correct answer is (A) \(1:4\).

Step by step solution

01

Write down the formula for root mean square velocity

The root mean square velocity (Vrms) of a gas is given by the formula: \(V_{rms} = \sqrt{\frac{3RT}{M}}\), where R is the universal gas constant, T is the temperature, and M is the molar mass of the gas.
02

Assign values to the constants and variables

For this exercise, let's assign the following values: R = 8.314 J/mol∙K (universal gas constant) T = temperature (constant for both Oxygen and Hydrogen) M_oxygen = M_o (molar mass of Oxygen) = 32 g/mol M_hydrogen = M_h (molar mass of Hydrogen) = 2 g/mol
03

Write down the Vrms equations for Oxygen and Hydrogen

Using the formula for root mean square velocity, we can write down the Vrms equations for Oxygen and Hydrogen: \(V_{rms,oxygen} = \sqrt{\frac{3RT}{M_o}}\) \(V_{rms,hydrogen} = \sqrt{\frac{3RT}{M_h}}\)
04

Write the velocity ratio

We want to find the ratio of the root mean square velocities of Oxygen and Hydrogen. We can write it as: \(\frac{V_{rms,oxygen}}{V_{rms,hydrogen}} = \frac{\sqrt{\frac{3RT}{M_o}}}{\sqrt{\frac{3RT}{M_h}}}\)
05

Simplify the velocity ratio

Now, let's simplify the ratio equation and solve for the velocity ratio: \(\frac{V_{rms,oxygen}}{V_{rms,hydrogen}} = \frac{\sqrt{\frac{3RT}{M_o}}}{\sqrt{\frac{3RT}{M_h}}} = \sqrt{\frac{M_h}{M_o}}\) Considering the molar masses of Oxygen and Hydrogen: \(\frac{V_{rms,oxygen}}{V_{rms,hydrogen}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}}\)
06

Calculate the velocity ratio

Now, let's calculate the root mean square velocity ratio: \(\frac{V_{rms,oxygen}}{V_{rms,hydrogen}} = \sqrt{\frac{1}{16}} = \frac{1}{4}\) So the correct ratio is \(1 : 4\). The correct answer is (A) \(1 : 4\).

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Most popular questions from this chapter

The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is (A) \(1: 8\) (B) \(1: 4\) (C) \(1: 16\) (D) \(1: 1\)

When the temperature of a gas is raised from \(27^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the percentage increase in the rms velocity of the molecules will be (A) \(15 \%\) (B) \(17.5 \%\) (C) \(10 \%\) (D) \(20 \%\)

At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its value at \(0^{\circ} \mathrm{C}\) ? (A) \(204.75 \mathrm{~K}\) (B) \(204.75^{\circ} \mathrm{C}\) (C) \(-204.75 \mathrm{~K}\) (D) \(-204.75^{\circ} \mathrm{C}\)

A gas at \(27{ }^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure $1 \mathrm{~s}$ allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (A) \(273^{\circ} \mathrm{C}\) (B) \(-173^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)

The volume of a gas at \(20 \mathrm{C}\) is \(200 \mathrm{ml}\). If the temperature is reduced to \(-20^{\circ} \mathrm{C}\) at constant pressure, its volume will be. (A) \(172.6 \mathrm{~m} 1\) (B) \(17.26 \mathrm{ml}\) (C) \(19.27 \mathrm{ml}\) (D) \(192.7 \mathrm{ml}\)
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