/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q30 Question: The mean free path of ... [FREE SOLUTION] | 91影视

91影视

Find study content
Learning Materials

Discover learning materials by subject, university or textbook.

Explanations

Textbooks
All Subjects

Anthropology

Archaeology

Architecture

Art and Design

Bengali

Biology

Business Studies

Greek

Chemistry

Chinese

Combined Science

Computer Science

Economics

Engineering

English

English Literature

Environmental Science

French

Geography

German

History

Hospitality and Tourism

Human Geography

Italian

Japanese

Law

Macroeconomics

Marketing

Math

Media Studies

Medicine

Microeconomics

Music

Nursing

Nutrition and Food Science

Physics

Polish

Politics

Psychology

Religious Studies

Sociology

Spanish

Sports Sciences

Translation

All Subjects

Biology

Business Studies

Chemistry

Combined Science

Computer Science

Economics

English

Environmental Science

Geography

History

Math

Physics

Psychology

Sociology
Features
Features

Discover all of these amazing features with a free account.

Flashcards

91影视 AI

Notes

Study Plans

Study Sets
What鈥檚 new?

Flashcards
Study your flashcards with three learning modes.

Study Sets
All of your learning materials stored in one place.

Notes
Create and edit notes or documents.

Study Plans
Organise your studies and prepare for exams.
91影视
Discover

All the hacks around your studies and career - in one place.

Find a degree

Magazine
Featured

Magazine
Trusted advice for anyone who wants to ace their studies & career.

The largest student job board with the most exciting opportunities.

Verified student deals from top brands.

Discover our mobile app to take your studies anywhere.

Learning Materials

Features

Discover

Chapter 19: Q30 (page 579)

Question: The mean free path of nitrogen molecules at 0.0⁰C and 1atmis $0.80 脳 10^{- 5} c m$ . At this temperature and pressure there are $2.7 脳 10^{19} m o l e c u l e s / c m^{3}$ . What is the molecular diameter?

Short Answer

Expert verified

Answer

The molecular diameter of nitrogen molecule is . 0.32 nm

Step by step solution

01

Given data

Atmospheric density

$\frac{N}{V} = 2.7 脳 10^{19} \frac{molecule}{{cm}^{3}} = 2.7 脳 10^{25} \frac{molecule}{m^{3}}$

Temperature;

$T = 0.0 掳 C$

Mean free path

$位 = 0.80 脳 10^{- 5} cm = 0.80 脳 10^{- 7} m$

Pressure

$P = 1.0 atm$

02

Understanding the concept

The expression for the mean free path is given by,

$位 = \frac{1}{\sqrt{2} 蟺 d^{2} (\frac{N}{V})}$

Here $位$ is the mean free path, $\frac{N}{V}$ is the molecule density in $m o l e c u l e / c m^{3}$ .

03

Calculate the molecular diameter

The mean free path is given by

$位 = \frac{1}{\sqrt{2} 蟺 d^{2} (\frac{N}{V})}$

On rearranging the equation in terms of d

$d = \sqrt{\frac{1}{\sqrt{2} 蟺位 (\frac{N}{V})}}$

Substitute $0.80 脳 10^{- 7} 尘蹿辞谤位$ and $2.7 脳 10^{25} \frac{molecule}{m^{3}} f o r \frac{位}{v}$ into the above equation

$d = \sqrt{\frac{1}{\sqrt{2} (3.142) (0.80 脳 10^{- 7}) (2.7 脳 10^{25})}} d = 3.22 脳 10^{- 10} m = 0.32 nm$

Therefore the molecular diameter of nitrogen molecule is 0.32 nm .

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Question: The speed of sound in different gases at a certain temperature T depends on the molar mass of the gases. Show that $\frac{v_{1}}{v_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$ whereis the speed of sound in a gas of molar mass M₁and v₂is the speed of sound in a gas of molar mass M₂.(Hint: See Problem 91.)

We know that for an adiabatic process, $p V^{纬} = a c o n s t a n t$ .Evaluate 鈥� constant鈥� for an adiabatic process involving exactly $2.0 鈥� m o l$ of an ideal gas passing through the state having exactly $p = 1.0 鈥� a t m$ and $T = 300 鈥� K$ . Assume a diatomic gas whose molecules rotate but do not oscillate

Question: For air near 0 ⁰C, by how much does the speed of sound increase for each increase in air temperature by 1 ⁰C? (Hint: See Problem 91)

Question: Air that initially occupies 0.140 m³at a gauge pressure of 103 kPais expanded isothermally to a pressure of 101.3 kPaand then cooled at constant pressure until it reaches its initial volume.

Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure).

The atmospheric density at an altitude of $2500 鈥� k m$ is about $1 m o l e c u l e / c m^{3}$ .

Assuming the molecular diameter of $2.0 脳 10^{- 8} c m$ , find the mean free path predicted by equation $位 = \frac{1}{\sqrt{2 鈥�} 蟺 d^{2} N / V}$
Explain whether the predicted value is meaningful.

See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Wave Optics

Read Explanation

Measurements

Read Explanation

Physical Quantities And Units

Read Explanation

Magnetism and Electromagnetic Induction

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.