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. 2.39

Expert-verified
An Introduction to Thermal Physics
Found in: Page 81
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

Compute the entropy of a mole of helium at room temperature and atmospheric pressure, pretending that all the atoms are distinguishable. Compare to the actual entropy, for indistinguishable atoms, computed in the text.

The actual entropy indistinguishable atoms is

See the step by step solution

Step by Step Solution

Step: 1 

The Sackur-Tetrode formula by ideal gas is

Where,represents volume, represents energy, represents the number of molecules, represents the mass of a single molecule, and represents Planck's constant.These are some of the assumptions used to generate this formula is that the molecules are indistinguishable, therefore altering any of the molecules makes no change in any arrangement of the molecules in position and momentum space. This assumption inserts the component into the multiplicity function's denominator.

The logarithm factor loses its , we get

Step: 2 Finding degree of freedom:

The mole mass of helium is ,the mass of helium molecule is

From ideal gas law, the pressure of and temperature of one mole occupies a volume of

The monatomic gas of internal energy is

Helium is monatomic gas so .

Step: 3 

By degree of freedom,

Substituting the values of , we get

Because there are many more molecular orbitals accessible to the system if the molecules are distinct, the entropy is substantially larger.

Most popular questions for Physics Textbooks

A black hole is a region of space where gravity is so strong that nothing, not even light, can escape. Throwing something into a black hole is therefore an irreversible process, at least in the everyday sense of the word. In fact, it is irreversible in the thermodynamic sense as well: Adding mass to a black hole increases the black hole's entropy. It turns out that there's no way to tell (at least from outside) what kind of matter has gone into making a black hole. Therefore, the entropy of a black hole must be greater than the entropy of any conceivable type of matter that could have been used to create it. Knowing this, it's not hard to estimate the entropy of a black hole.
Use dimensional analysis to show that a black hole of mass should have a radius of order , where is Newton's gravitational constant and is the speed of light. Calculate the approximate radius of a one-solar-mass black hole . In the spirit of Problem , explain why the entropy of a black hole, in fundamental units, should be of the order of the maximum number of particles that could have been used to make it.

To make a black hole out of the maximum possible number of particles, you should use particles with the lowest possible energy: long-wavelength photons (or other massless particles). But the wavelength can't be any longer than the size of the black hole. By setting the total energy of the photons equal to , estimate the maximum number of photons that could be used to make a black hole of mass . Aside from a factor of , your result should agree with the exact formula for the entropy of a black hole, obtained* through a much more difficult calculation:

Calculate the entropy of a one-solar-mass black hole, and comment on the result.

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.