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Expert-verified Found in: Page 97 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # 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 $S=-\frac{3}{2}NK\mathrm{ln}\left(\frac{2U}{3NK}\right)+f\left(N,V\right)$ and the graph can be sketched as below. See the step by step solution

## Step 1: Given Information

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

${C}_{V}=-\frac{3}{2}NK$

Where,

$N$ 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:

$S=\int \frac{{C}_{V}\left(T\right)}{T}dT$

Where,

${C}_{V}$ = specific heat

$T$ = Temperature in Kelvin

By substituting the value of ${C}_{V}$ in the above equation, we get,

$S=\int -\frac{3}{2}\left(\frac{NK}{T}\right)dT\phantom{\rule{0ex}{0ex}}S=-\frac{3}{2}NK\int \left(\frac{1}{T}\right)dT\phantom{\rule{0ex}{0ex}}S=-\frac{3}{2}NKT\mathrm{ln}\left(T\right)+f\left(N,V\right)..........\left(1\right)$

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:

$U=-K=-\frac{3}{2}NKT$

By rearranging the terms, we get,

$T=-\frac{2U}{3NK}$

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

$S=-\frac{3}{2}NK\mathrm{ln}\left(\frac{2U}{3NK}\right)+f\left(N,V\right)$

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

$S=-\frac{3}{2}NK\mathrm{ln}\left(U\right)-\frac{3}{2}NK\mathrm{ln}\left(\frac{3NK}{2}\right)+f\left(N,V\right)\phantom{\rule{0ex}{0ex}}S=-\frac{3}{2}NK\mathrm{ln}\left(U\right)+g\left(N,V\right)$

From the above equation, the graph can be plotted as below: Hence, the required expression is: $S=-\frac{3}{2}NK\mathrm{ln}\left(\frac{2U}{3NK}\right)+f\left(N,V\right)$

The graph of entropy as a function can be sketched as follow:  ### Want to see more solutions like these? 