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Question: Do you expect the entropy of an interacting gas of N particles in volume V at temperature T to be smaller or bigger than the entropy for an ideal gas of the same (N, V, T)?

My understanding could be flawed. If so, I am happy to corrected.

My guess is that the entropy of the nonideal gas should be greater. This is because nonideal processes are irreversible and by the second law of thermodynamics we have to factor in an increase in entropy of the universe. Whereas in the ideal gas situation, we don't have to factor this in.

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    $\begingroup$ "Ideal" means different things in different contexts. For a gas, "ideal" means that no interaction exist except for collision of point particles. For a process, "ideal" can mean that no gradients exist and that no entropy is generated. A gas is not a process. $\endgroup$ – Chemomechanics Apr 28 '17 at 17:12
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    $\begingroup$ Think of what you mean by "interaction". Will the particles clump together in some way (attractive force, reducing the multiplicity) or repel each other, (increasing the number of microstates and entropy) $\endgroup$ – user154420 Apr 28 '17 at 19:48
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Consider the definition of entropy in the microcanonical ensemble:

$$S=k_B \log \Omega(N,V,E)$$

where

$$\Omega(N,V,E)=\frac 1 {N! h^{3N}} \int_{\mathcal H<E} d^N\mathbf p \ d^N \mathbf q \tag{1}\label{1}$$

is the microcanonical partition function. $\mathcal H$ is the hamiltionian of the system, which depends on all the momenta $\mathbf p$ and all the generalized coordinates $\mathbf q$, and the energy of the system is $E$.

The entropy is therefore proportional to the logarithm of the hypervolume delimited by the energy hypersurface $\mathcal H = E$.

The hamiltionian can be written in general as the sum of a kinetic term depending on the momenta and an interaction term:

$$\mathcal H (\{\mathbf p, \mathbf q\})=\sum_{i=1}^N\frac{p_i^2}{2 m}+U(\{\mathbf q\})$$

For an ideal gas, the interaction term is zero, so that the energy hypersurface is an hypersphere:

$$\sum_{i=1}^N p_i^2=2 m E$$

The integral in Eq.\ref{1} becomes in this case trivial and can be easily solved, holding the famous Sackur-Tetrode expression for the entropy of an ideal gas:

$$S_{ig}=Nk_B \left [\log \left(\frac V N u^{3/2}\right) +\frac 3 2 \left( \frac 5 3 + \log \frac{4 \pi m}{3h^2}\right) \right]\tag{2}\label{2}$$

where $u=E/N$.

Your question therefore boils down to the following: is

$$\int_{\sum_{i=1}^N p_i^2/2 m+U(\{\mathbf q\}) <E} d^N\mathbf p \ d^N \mathbf q $$

smaller or bigger than

$$\int_{\sum_{i=1}^N p_i^2/2 m <E} d^N\mathbf p \ d^N \mathbf q \ \ ?$$

At first sight, it would seem that it all depends on the interaction term $U(\{\mathbf q\})$. Let's consider the very simple case of a single particle in 1D confined in a segment of length $L$(*); in this case, for an ideal gas ($U=0$) we simply have to calculate the area of a rectangle:

$$\Omega_{ig} = \frac 1 h \int_0^L dq \int_{|p|<\sqrt{2mE}} dp = \frac 1 h 2L \sqrt{2mE} $$

let's now add a term dependent on $q$ by defining

$$\mathcal H(p,q) = \frac{p^2}{2m} + \frac 1 2 m \omega^2 q^2$$

(you will have noticed that this is the hamiltonian of an harmonic oscillator). The hypersurface is an ellipse:

$$\frac{p^2}{2mE} + \frac 1 {2E} m \omega^2 q^2 = \frac{p^2}{a^2} + \frac{q^2}{b^2} = 1$$

and the axes of the ellipse are $a=\sqrt{2mE}$ and $b=\sqrt{2E/m\omega^2}$. If $2b<L$, the partition function is simply the area of the ellipse divided by $h$:

$$\Omega = \frac 1 h A = \frac 1 h \pi a b = \frac {2 \pi E}{h \omega}$$

If $2b>L$, however, it will be the area of the intersection between the ellipse and the rectangle of the ideal gas. A graphical representation could help to understand:

enter image description here

You can see that no matter the value of $b$, you will always get an area which is smaller than the area of the rectangle corresponding to $\Omega_{ig}$ (**). You can try to devise different forms for the interaction term, but the result will always be the same: you won't be able to obtain an area larger than the one obtained for an ideal gas.

This is because an interaction term always introduce an ulterior limitation to the domain of the integral of Eq.\ref{1}, resulting always in a smaller hypervolume in phase space. We can therefore conclude that, at given $N,V$ and $E$ the entropy of an ideal gas is the largest possible:

$$S \leq S_{ig}$$

This analysis was done in the microcanonical ensemble. However, if $N$ is large, fixing the temperature (canonical ensemble) is equivalent to fixing the energy (microcanonical ensemble), so the result also holds for the canonical ensemble. You could try in principle to show it with the formalism of the canonical ensemble, but it would be slightly more complicated.


(*) In this case we should not use definition (1) of the partition function, since that definition is actually an approximation for large $N$ (see K.Huang, Statistical Mechanics). However, I just want to prove a point here, and taking a $1D$ example is the best way to do so.

(**) Except for $b \to \infty$.

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  • $\begingroup$ I cannot pin down where I disagree in your formalism , (evidently you are one of those who down voted my opposite statement), but as a particle physicist used to thinking with feynman diagrams for microstates, it seems incredible that the number of microstates of an ideal gas, kinematic contacts, are larger that the number of microstates of a real gas, where there are a number of extra interactions generating microstates for the system. I disagree but I am not downvoting because I cannot pin point where your mathematics goes off. $\endgroup$ – anna v Jan 2 '18 at 13:23
  • $\begingroup$ OK, I think I see that you are demonstrating the statement that the classical ideal gas has an infinity of microstates because the variables are not quantized but continuous. You are mixing the classical with the underlying quantum states. $\endgroup$ – anna v Jan 2 '18 at 13:28
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    $\begingroup$ No, this has nothing to do with quantum mechanics. Also you do not obtain an infinite entropy for the ideal gas: I also gave you the expression ($S_{ig}$), and the result is not infinite. As I wrote, an ideal gas has higher entropy because interactions actually reduce the accessible phase space volume. Maybe you are thinking of additional internal degrees of freedom? Those would actually increase the phase space volume and therefore the entropy, but in a different way. Maybe I can add a comment about this... $\endgroup$ – valerio Jan 2 '18 at 14:22
  • $\begingroup$ I am thinking of all the possible degrees of freedom introduced by the quantum mechanical states. The list of interactions in my answer has extra degrees of freedom, as that is what quantum mechanical potentials in molecules are about. The infinity for the classical case is a quote, not my idea, physics.umd.edu/courses/Phys603/kelly/Notes/Semiclassical.pdf . $\endgroup$ – anna v Jan 2 '18 at 15:15
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    $\begingroup$ This problem only concerns the definition of a microstate. If the energy of the system is finite, then the integral in (1) is finite, and therefore the entropy is finite and given by equation (2). $\endgroup$ – valerio Jan 3 '18 at 8:12
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$ S = k_{B} \ln \Omega \; $ where $\Omega $ is the number of accessible microstates.

$\Omega$ can be characterized by $(q, p)$ the position $q$ and momentum $p$ of particles.

"Naive" physical intuition:

Since interacting gas (e.g. van der Waal's gas) tends to attract each other, these gas particles cannot be separated too far away, compared with ideal gas.

Hence, I guess $\Omega_{non-ideal} < \Omega_{ideal} \implies S_{non-ideal} < S_{ideal}$.

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    $\begingroup$ Gas particles can be separated far away, that's why they are called gases in the first place. You should consider revising your argument and make it more rigorous. $\endgroup$ – Deep Apr 28 '17 at 6:29
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Wikipedia has a list of extra interactions in real ( non ideal gases) .

compressibility effects;

variable specific heat capacity;

van der Waals forces;

non-equilibrium thermodynamic effects;

issues with molecular dissociation and elementary reactions with variable composition

Classical physics , including thermodynamics, emerges from the underlying quantum mechanical level.

For a given macroscopic system, there are many microscopic states. A key idea from quantum mechanics is that the states of atoms, molecules, and entire systems are discretely quantized. This means that a system of particles under certain constraints, like being in a box of a specified size, or having a fixed total energy, can exist in a finite number of allowed microscopic states. This number can be very big, but it is finite. The microstates of the system keep changing with time from one quantum state to another as molecules move and collide with one another. The probability for the system to be in a particular quantum state is defined by its quantum-state probability $ p_i$ . The set of the $ p_i$ is called the distribution of probability. The sum of the probabilities of all the allowed quantum states must be unity, hence for any time $ t$ ,

$\displaystyle \sum_i p_i = 1.$

Using the definition of entropy,

entropy

where $p_i$ is the probability for the gas to be in a microstate i and k is the Boltzman constant. (Because the $ p_i$ are less than unity, the constant is chosen to be negative to make the entropy positive. )

One sees that the more microstates the larger the entropy.

Since a non ideal gas has many more interactions between the particles that compose it, there are many more microstates than for an ideal gas.

Thus a real gas for the same temperature and pressure will have higher entropy.

Edit after downvotes:

One should not mix two frameworks, the classical thermodynamics one and the underlying quantum mechanical one. In my case I am also guilty because I did not define that the "ideal gas" in the quantum mechanical frame, means the particles are at such a temperature that no quantum states can be excited, or a very small percent ( due to the black body distribution from the tails there will always be some energy to excite molecular states). That will be the minimum microstate, and minimum energy.

In the classical thermodynamics case, the form of entropy as microstates is meaningless, as, because the variables are continuous,then number of microstates and thus will be infinite, giving a maximum for entropy.

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