• :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. 3.15

An Introduction to Thermal Physics
Found in: Page 97
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.


Short Answer

In Problem 1.55 you used the virial theorem to estimate the heat capacity of a star. Starting with that result, calculate the entropy of a star, first in terms of its average temperature and then in terms of its total energy. Sketch the entropy as a function of energy, and comment on the shape of the graph.

The required expression for the entropy of a star is S=-32NKln2U3NK+f(N,V) and the graph can be sketched as below.

See the step by step solution

Step by Step Solution

Step 1: Given Information

The heat capacity of a star that was estimated using the virial theorem is given as:



N is the number of particles (typically dissociated protons and electrons).

The negative sign symbolizes that it is a gravitational bound system.

Step 2: Calculation

The change in entropy is given as:



CV = specific heat

T = Temperature in Kelvin

By substituting the value of CV in the above equation, we get,

S=-32NKTdTS=-32NK1TdTS=-32NKTln(T)+f(N,V) ..........(1)

In this equation, f is the function of N and volume V.

Total energy of gravitationally bound system is negative and from the virial theorem, it is found that:


By rearranging the terms, we get,


By substituting this value in equation (1), we get,


For plotting the graph, let us further simplify the above equation,


From the above equation, the graph can be plotted as below:

Step 3: Final answer

Hence, the required expression is: S=-32NKln2U3NK+f(N,V)

The graph of entropy as a function can be sketched as follow:

Most popular questions for Physics Textbooks


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.