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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

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) \(400 \mathrm{~ms}^{-1}\)

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) \(100^{\circ} \mathrm{C}\)

Calculate the temperature at which \(\mathrm{rms}\) velocity of \(\mathrm{SO}_{2}\) molecules is the same as that of \(\mathrm{O}_{2}\) molecules at \(27^{\circ} \mathrm{C}\). Molecular weights of Oxygen and \(\mathrm{SO}_{2}\) are \(32 \mathrm{~g}\) and \(64 \mathrm{~g}\) respectively (A) \(327^{\circ} \mathrm{C}\) (B) \(327 \mathrm{~K}\) (C) \(127^{\circ} \mathrm{C}\) (D) \(227^{\circ} \mathrm{C}\)

The specific heats at constant pressure is greater than that of the same gas at constant volume because (A) At constant volume work is done in expanding the gas. (B) At constant pressure work is done in expanding the gas. (C) The molecular vibration increases more at constant pressure. (D) The molecular attraction increases more at constant pressure.

The rms speed of a gas at a certain temperature is \(\sqrt{2}\) times than that of the Oxygen molecule at that temperature, the gas is (A) \(\mathrm{SO}_{2}\) (B) \(\mathrm{CH}_{4}\) (C) \(\mathrm{H}_{2}\) (D) \(\mathrm{He}\)
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