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Chapter 9: Problem 1300

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) \(75 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The new rms velocity of the gas molecules with twice the molecular weight and half the absolute temperature is (C) \(600 \mathrm{~ms}^{-1}\).

Step by step solution

01

Write the formula for rms velocity

The formula for the rms velocity of gas molecules is \(v_\text{rms} = \sqrt{\frac{3kT}{m}}\)
02

Write the given rms velocity, molecular weight, and temperature

The initial rms velocity is \(300 \text{ m/s}\), the molecular weight is represented by \(m\), and the initial temperature is represented by \(T\). We want to find the rms velocity of gas molecules with twice the molecular weight (i.e., \(2m\)) and half the absolute temperature (i.e., \(\frac{1}{2}T\)).
03

Find the ratio of the new rms velocity to the initial rms velocity

Divide the formula of the new rms velocity by the initial rms velocity: \(\frac{v_\text{new}}{v_\text{rms}} = \frac{\sqrt{\frac{3k(\frac{1}{2}T)}{2m}}}{\sqrt{\frac{3kT}{m}}}\)
04

Simplify the ratio

Now, we will simplify the expression: \(\frac{v_\text{new}}{v_\text{rms}} = \sqrt{\frac{\frac{3k(\frac{1}{2}T)}{2m}}{\frac{3kT}{m}}}\) \(\frac{v_\text{new}}{v_\text{rms}} = \sqrt{\frac{\frac{3kT}{4m}}{\frac{3kT}{m}}}\) \(\frac{v_\text{new}}{v_\text{rms}} = \sqrt{\frac{4m}{m}}\) \(\frac{v_\text{new}}{v_\text{rms}} = \sqrt{4}\) \(\frac{v_\text{new}}{v_\text{rms}} = 2\)
05

Calculate the new rms velocity

Now that we have the ratio, we can find the new rms velocity: \(v_\text{new} = 2 \times v_\text{rms}\) \(v_\text{new} = 2 \times 300 \text{ m/s}\) \(v_\text{new} = 600 \text{ m/s}\)
06

Match the result with the choices

The new rms velocity of the gas molecules is \(600 \text{ m/s}\), which matches choice (C). So, the correct answer is (C) \(600 \mathrm{~ms}^{-1}\).

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Most popular questions from this chapter

\(2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas is taken at \(27^{\circ} \mathrm{C}\) and pressure \(76 \mathrm{~mm} \mathrm{Hg}\). Find out volume of gas (ln liter) (A) \(3.08\) (B) \(44.2\) (C) \(2.05\) (D) \(2.44\)

If three molecules have velocities \(0.5,1\) and 2 the ratio of rms speed and average speed is (The velocities are in \(\mathrm{km} / \mathrm{s}\) ) (A) \(0.134\) (B) \(1.34\) (C) \(1.134\) (D) \(13.4\)

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 rms speed of the molecules of a gas at a pressure \(10^{5} \mathrm{~Pa}\) and temperature \(0^{\circ} \mathrm{C}\) is \(0.5 \mathrm{~km} / \mathrm{s}\). If the pressure is kept constant but temperature is raised to $819^{\circ} \mathrm{C}$, the rms speed becomes (A) \(1.5 \mathrm{kms}^{-1}\) (B) \(2 \mathrm{kms}^{-1}\) (C) \(1 \mathrm{kms}^{-1}\) (D) \(5 \mathrm{kms}^{-1}\)

For a gas, the rms speed at \(800 \mathrm{~K}\) is (A) Four times the value at \(200 \mathrm{~K}\) (B) Twice the value at \(200 \mathrm{~K}\) (C) Half the value at \(200 \mathrm{~K}\) (D) same as at \(200 \mathrm{~K}\)
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