Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

Use the formula $P = - (\partial U / \partial V)_{S, N}$ to show that the pressure of a photon gas is 1/3 times the energy density (U/V). Compute the pressure exerted by the radiation inside a kiln at 1500 K, and compare to the ordinary gas pressure exerted by the air. Then compute the pressure of the radiation at the centre of the sun, where the temperature is 15 million K. Compare to the gas pressure of the ionised hydrogen, whose density is approximately 10⁵ kg/m³.

Hence, the pressure exerted inside a kiln is $P = 1.272 \times 10^{- 3} Pa$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

Formula to be used is $P = - (\partial U / \partial V)_{S, N}$

Step 2: Explanation

At constant N and S, the pressure equals the negative of the partial derivative of the total energy with respect to the volume, i.e.

$P = - {(\frac{\partial U}{\partial V})}_{N, S} (1)$

To calculate the pressure, we must express the total energy in terms of entropy. The total energy is calculated as follows:

$U = \frac{8 π^{5} (k T)^{4}}{15 (h c)^{3}} V$

The entropy is given as:

$S = \frac{32 π^{5}}{45} V {(\frac{k T}{h c})}^{3} k$

Let

$A = \frac{8 π^{2} k^{4}}{15 (h c)^{3}}$

Total energy and entropy is:

$U = A V T^{4} and S = \frac{4}{3} A V T^{3}$

Solve entropy equation for T:

$T = {(\frac{3 S}{4 A V})}^{1 / 3}$

Substitute T in total energy:

$U = A V {(\frac{3 S}{4 A V})}^{4 / 3} U = A {(\frac{3 S}{4 A})}^{4 / 3} \frac{1}{V^{1 / 3}}$

Substitute in equation (1)

$P = - A {(\frac{3 S}{4 A})}^{4 / 3} \frac{\partial}{\partial V} (\frac{1}{V^{1 / 3}}) P = - A {(\frac{3 S}{4 A})}^{4 / 3} (- \frac{1}{3} \frac{1}{V^{4 / 3}}) P = \frac{1}{3} A {(\frac{3 S}{4 A V})}^{4 / 3}$

But,

$T = {(\frac{3 S}{4 A V})}^{1 / 3}$

Therefore,

$P = \frac{1}{3} A T^{4}$

Step 3: Explanation

The total energy of radiation is:

$U = \frac{8 π^{5} (k T)^{4}}{15 (h c)^{3}} V$

Substitute the values:

$\frac{U}{V} = \frac{8 π^{5} {((1.38 \times 10^{- 23} J / K) (1500 K))}^{4}}{15 {((6.626 \times 10^{- 34} J \cdot s) (3.0 \times 10^{8} m / s))}^{3}} = 3.815 \times 10^{- 3} J / m^{3}$

The pressure is:

$P = \frac{1}{3} (3.815 \times 10^{- 3} J / m^{3}) = 1.272 \times 10^{- 3} J / m^{3} = 1.272 \times 10^{- 3} Pa P = 1.272 \times 10^{- 3} Pa$

For comparison, the pressure within the kiln is the same as the pressure outside it, which is 1 atm in pascal 1.01 x 10⁵ Pa, which is about 10⁸ higher than the pressure caused by photons. The temperature at the sun's core is T = 15 x 10⁶ K, hence the energy per unit volume is:

$\frac{U}{V} = \frac{8 π^{5} {((1.38 \times 10^{- 23} J / K) (15 \times 10^{6} K))}^{4}}{15 {((6.626 \times 10^{- 34} J \cdot s) (3.0 \times 10^{8} m / s))}^{3}} = 3.815 \times 10^{13} J / m^{3}$

The pressure is:

role="math" localid="1647764290491" $P = \frac{1}{3} (3.815 \times 10^{13} J / m^{3}) = 1.272 \times 10^{13} J / m^{3} = 1.272 \times 10^{13} Pa P = 1.272 \times 10^{13} Pa$

Number of moles per unit volume equals to:

$\frac{n}{V} = (10^{3} mole / kg) (10^{5} kg / m^{3}) = 10^{8} mole / m^{3}$

The pressure is:

$P = \frac{n R T}{V} = 2 (10^{8} mole / m^{3}) (8.314 J / K \cdot mole) (15 \times 10^{6} K) = 2.5 \times 10^{16} Pa$

The factor 2 came from the fact that each ionised hydrogen atom has and electron and a proton.

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