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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

At what temperature is the kinetic energy of a gas molecule double that of its value at \(27^{\circ} \mathrm{C}\) (A) \(54^{\circ} \mathrm{C}\) (B) \(108^{\circ} \mathrm{C}\) (C) \(327^{\circ} \mathrm{C}\) (D) \(300^{\circ} \mathrm{C}\)

The speeds of 5 molecules of a gas (in arbitrary units) are as follows: \(2,3,4,5,6\), The root mean square speed for these molecules is (A) \(4.24\) (B) \(2.91\) (C) \(4.0\) (D) \(3.52\)

A sealed container with negligible co-efficient of volumetric expansion contains helium (a monatomic gas) when it is heated from \(200 \mathrm{~K}\) to \(400 \mathrm{~K}\), the average \(\mathrm{K} . \mathrm{E}\). of helium atom is (A) Halved (B) Doubled (C) Unchanged (D) Increased by factor \(\sqrt{2}\)

If \(\mathrm{rms}\) speed of a gas is $\mathrm{v}_{\mathrm{rms}}=1840 \mathrm{~m} / \mathrm{s}\( and its density \)\rho=8.99 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{3}$, the pressure of the gas will be (A) \(1.01 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(1.01 \times 10^{7} \mathrm{Nm}^{-2}\) (D) \(1.01 \mathrm{Nm}^{-2}\)

A vessel contains 1 mole of \(O_{2}\) gas (relative molar mass 32 ) at a temperature \(\mathrm{T}\). The pressure of the gas is \(\mathrm{P}\). An identical vessel containing 1 mole of He gas (relative molar mass 4 ) at a temperature \(2 \mathrm{~T}\) has pressure of \(.\) (A) \(8 \mathrm{P}\) (B) \((\mathrm{p} / 8)\) (C) \(2 \mathrm{P}\) (D) \(\mathrm{P}\)
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