• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q 6.14

Expert-verified
An Introduction to Thermal Physics
Found in: Page 228
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere, already derived in Problems 1.16 and 3.37. (Hint: Let the system be a single air molecule, let s1 be a state with the molecule at sea level, and let s2 be a state with the molecule at height z.)

Therefore, the exponential formula for the density of an isothermal atmosphere is:ρ(z)=ρ(0)e-mgz/kT

See the step by step solution

Step by Step Solution

Step 1: Given information

Let the system be a single air molecule, let S1 be a state with the molecule at sea level, and let S2 be a state with the molecule at height z.

Step 2: Explanation

Consider a system with a single air molecule, where S1 is the state when the molecule is at sea level and S2 is the state when the molecule is at a height of 2. Assume that the energy is only potential energy, so the difference in energy between the states S1 and S2 is the potential energy, which is ΔE=mgz and the ratio of S2state probability to state s1 probability is:

role="math" localid="1647369610229" Ps2Ps1=e-E2/kTe-E1/kT=e-ΔE/kTPs2Ps1=e-mgz/kT

This means that the air molecule is less likely to be at height of z than at the see level by a factor of e-mgz/kT, and that the number of molecules per unit volume at height of z is also smaller than the see level by the same ratio in the isothermal atmosphere, so:

ρ(z)=ρ(0)e-mgz/kT

Most popular questions for Physics Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.