Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 147

At what temperature will He atoms have the same \(u_{\mathrm{rms}}\) value as \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The temperature at which He atoms will have the same \(u_{rms}\) value as \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C}\) is \(2,725.96^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate \(u_{rms}\) for \(\mathrm{N}_{2}\)

First, calculate the root-mean-square speed for N2 using the formula \(u_{rms} = \sqrt{\frac{3kT}{m}}\), with \(T = 25^{\circ}C = 298.16K\), \(k = 1.38x10^{-23} \,J/K-atom\), and \(m = 28.01 \,g/mol, m = \frac{28.01 \times 10^{-3}}{6.02 \times 10^{23}} kg/atom\)
02

Use \(\mathrm{N}_{2}\) \(u_{rms}\) to calculate temperature for He

Now use this \(u_{rms}\) value and the molar mass of helium (4.003 g/mol, so \(m = \frac{4.003 \times 10^{-3}}{6.02 \times 10^{23}} kg/atom\)) in the root-mean-square speed equation and solve for \(T = \frac{m u_{rms}^{2}}{3k}\)
03

Convert the temperature to Celsius

The temperature calculated in Step 2 will be in Kelvin. Convert it to Celsius using the relation \(C = K - 273.16\).

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

In 1995 a man suffocated as he walked by an abandoned mine in England. At that moment there was a sharp drop in atmospheric pressure due to a change in the weather. Suggest what might have caused the man's death.

A certain anesthetic contains 64.9 percent \(\mathrm{C}, 13.5\) percent \(\mathrm{H},\) and 21.6 percent \(\mathrm{O}\) by mass. At \(120^{\circ} \mathrm{C}\) and \(750 \mathrm{mmHg}, 1.00 \mathrm{~L}\) of the gaseous compound weighs \(2.30 \mathrm{~g}\). What is the molecular formula of the compound?

A mixture of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{MgCO}_{3}\) of mass \(7.63 \mathrm{~g}\) is reacted with an excess of hydrochloric acid. The \(\mathrm{CO}_{2}\) gas generated occupies a volume of \(1.67 \mathrm{~L}\) at 1.24 atm and \(26^{\circ} \mathrm{C}\). From these data, calculate the percent composition by mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in the mixture.

In 2012 , Felix Baumgartner jumped from a balloon roughly \(24 \mathrm{mi}\) above Earth, breaking the record for the highest skydive. He reached speeds of more than 700 miles per hour and became the first skydiver to exceed the speed of sound during free fall. The helium-filled plastic balloon used to carry Baumgartner to the edge of space was designed to expand to \(8.5 \times 10^{8} \mathrm{~L}\) in order to accommodate the low pressures at the altitude required to break the record. (a) Calculate the mass of helium in the balloon from the conditions at the time of the jump \((8.5 \times\) \(\left.10^{8} \mathrm{~L},-67.8^{\circ} \mathrm{C}, 0.027 \mathrm{mmHg}\right) .\) (b) Determine the volume of the helium in the balloon just before it was released, assuming a pressure of 1.0 atm and a temperature of \(23^{\circ} \mathrm{C}\).

Assuming that air contains 78 percent \(\mathrm{N}_{2}, 21\) percent \(\mathrm{O}_{2}\), and 1 percent Ar, all by volume, how many molecules of each type of gas are present in \(1.0 \mathrm{~L}\) of air at STP?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

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