What is the Issue with Extensive Properties of Entropy?

  • Thread starter Thread starter Vectorcrust
  • Start date Start date
AI Thread Summary
The discussion revolves around the extensive property of entropy and its behavior when the number of particles, volume, and energy of an ideal gas are doubled. A participant initially expresses confusion about why entropy does not simply double under these conditions. Another contributor clarifies that while the number of particles affects the entropy logarithmically, the doubling of energy and volume introduces positive terms that cancel out the negative term from the particle count. The original poster acknowledges the misunderstanding and appreciates the clarification. The conversation highlights the complexities of statistical mechanics and the specific application of the entropy formula to monoatomic gases.
Vectorcrust
Messages
5
Reaction score
0
Hi to all!

The entropy is known as an extensive property. Here is an expression for the entropy of ideal gas derived by statistical mechanics methods:

sgas2.gif


Imagine that I multiply by 2 the number of particles, the volume of particles and the energy of particles(so the molar volume and molar energy and all other intensive properties are the same). According to this expression I'll never get entropy multiplied by 2 because of N that under ln expression.

Where am I wrong?

Thanks.
 
Last edited:
Physics news on Phys.org
Hi Vectorcrust, welcome to PF!

Vectorcrust said:
According to this expression I'll never get entropy multiplied by 2 because of N that under ln expression.

Sure you will, if you note that U and V get doubled as well and you work through the logarithms.
 
The U and V logarithms will give you postive ln(2) terms and the logarithm with N inside will give you -ln(2) terms. They will cancel.
 
Mapes said:
Sure you will, if you note that U and V get doubled as well and you work through the logarithms.


I see it now. Sorry for dumb question. Thanks a lot!
 
Vectorcrust said:
I see it now. Sorry for dumb question. Thanks a lot!

Not dumb! Stat. mech. requires a lot of staring at logarithm-filled equations no matter how smart you are. Stick around!
 
also note that the formula you posted is not for any "ideal gas", it is for a monoatomic gas. Just keep that in mind.
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top