Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 1297

The root mean square velocity of a gas molecule of mass \(\mathrm{m}\) at a given temperature is proportional to (A) \(\mathrm{m}^{0}\) (B) \(\mathrm{m}^{-1 / 2}\) (C) \(\mathrm{m}^{1 / 2}\) (D) \(\mathrm{m}\)

Short Answer

Expert verified
The root mean square velocity of a gas molecule with mass m at a given temperature is proportional to \(m^{-1/2}\), as described by the formula \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(v_{rms}\) is the root mean square velocity, \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of the gas molecule. Comparing this expression to the given options, we find the correct answer is Option (B): \(v_{rms} \propto m^{-1/2}\).

Step by step solution

01

Root Mean Square Velocity Formula

The root mean square velocity \(v_{rms}\) of a gas molecule can be found using the kinetic theory of gases, which relates the molecular mass, temperature, and heat capacity: \[v_{rms} = \sqrt{\frac{3kT}{m}}\] Here, \(v_{rms}\) = root mean square velocity \(k\) = Boltzmann constant \(T\) = temperature \(m\) = mass of the gas molecule
02

Find Proportionality

Now, let's find the proportionality between the mass of the gas molecule and the root mean square velocity. From the formula, we have: \[v_{rms} \propto \sqrt{\frac{3kT}{m}}\] Since \(3kT\) is a constant term, we can write: \[v_{rms} \propto \frac{1}{\sqrt{m}}\] or equivalently \[v_{rms} \propto m^{-1/2}\] Comparing this expression to the given options, we can see that it matches Option (B): \[v_{rms} \propto m^{-1/2}\] Thus, the root mean square velocity of a gas molecule with mass m at a given temperature is proportional to \(m^{-1/2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 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) \(\mathrm{He}\)

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

\(\mathrm{O}_{2}\) gas is filled in a vessel. If pressure is double, temperature becomes four times, how many times its density will become. (A) 4 (B) \((1 / 4)\) (C) 2 (D) \((1 / 2)\)

The root mean square speed of hydrogen molecules at \(300 \mathrm{~K}\) is $1930 \mathrm{~m} / \mathrm{s}$. Then the root mean square speed of Oxygen molecules at \(900 \mathrm{~K}\) will be (A) \(836 \mathrm{~m} / \mathrm{s}\) (B) \(643 \mathrm{~m} / \mathrm{s}\) (C) \(1930 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (D) \([(1930) / \sqrt{3}] \mathrm{m} / \mathrm{s}\)
See all solutions

Recommended explanations on English Textbooks

Free Response Essay

Read Explanation

History of English Language

Read Explanation

Morphology

Read Explanation

Textual Analysis

Read Explanation

Semiotics

Read Explanation

Sociolinguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free