# Tag Info

22

While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically. In deriving the VDW equation, the particles are still assumed to be hard spheres, but this assumption is corrected for with the parameter $a$. The hard sphere ...

20

The differences in acceleration due to gravity is not the main factor in comparing how accurate the approximation is for each planet. The main factor is the mass of gas each planet's atmosphere contains. Mercury has almost no atmosphere. The total mass of all gas in Mercury's atmosphere is only 10000 kg! The pressure is less than $10^{-14}$ bar. The ...

10

You have to bring both equations into the same form and then compare the coefficients. You seem not to have used the tip with the Taylor expansion and I guess that's why you got an incorrect result. As you wrote: The virial equation of state is \begin{align} p &=\frac{nRT}{V} \left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+ \ldots \right) \\ &= \frac{RT}{...

9

The authors attempt to explain that in many sources the term "ideal gas" is used in place of "perfect gas" to indicate a gas following the ideal gas law and which has the property that the molecules do not interact. They equate the absence of an intermolecular interaction with the statement that the "interactions are zero". In other words, there is no ...

8

The short answer to your question is because that is what the experimental data showed. The constants in the van der Waals equation were originally derived empirically for each gas. The constant $b$ happens to correlate well with four times the molecular volume for a number of gasses, which lets us calculate it for other gasses. To answer your question a ...

8

The trend you've noticed here is a very interesting way of studying the interactions of a gas with itself. You are certainly right about how to find the minimum of these curves, but the important thing here is what that minimum actually means. $$Z=\frac{PV}{RT}$$By analyzing this, we expect that $Z$ will tell us something useful about the way the volume of ...

7

Pressure correction term depends upon: Number of molecules attracting the molecules which comes to strike the wall and as such it is proportional to density of gas i.e. proportional to $n/V$ where $n$ is the number of moles of gas and $V$ is the volume of the container. It also depends upon number of molecules which has a strike the unit area of the wall ...

7

Summing up the argument from the Wikipedia derivation, the reason it is proportional to $(\frac{1}{V_\mathrm m})^2$ where $V_\mathrm m=\frac Vn$ (the molar volume) comes from how the molecules experience the attractive force. The only molecules that experience a net attractive force are those close to the edge of the container because as we move toward the ...

6

You need to use the equation $$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$

5

It is true that a high order polynomial can fit any training set. But that is not a strength - an unfalsifiable model overfits. In particular, a polynomial of order n is only likely to be predictive if the true function is n times differentiable. Since chemical space is discrete, and for many purposes some molecules are special cases, polynomial models are a ...

5

For n data points there exists an n-1 order polynomial that perfectly fits the data. Therefore there is no basis for "a neural network or whatever" being better. Furthermore, it simply isn't true that the Peng-Robinson equation "has no underlying physical meaning". The Peng-Robinson equation (like Van der Waals) recognizes that atoms/molecules occupy ...

5

In general no, the pressure will not be the same, although there will be intersection points where (by coincidence) they will be the same, and there will be regions of the phase diagram for some equations of state (EOS) and for some substances that might be close enough that it doesn't matter. For example, the van der Waals EOS gives pressure values at high ...

5

You are right in saying that in your question, the gas is an non-ideal (real) gas. Basically, the reason why we use the ideal gas equation despite knowing that it is a real gas is simply because that the equation for real gases is too complicated and contain more variables which are hard to find given limited data. The reason for this complicated formula is ...

5

As pointed out already, the van der Waals (vdW) equation is also an empirical equation. However, I think what you are saying is that the vdW equation is a physically motivated empirical equations where the empirical parameters in principle correspond to something "real" (e.g. molecular size and interaction strength) in contrast to a fit to an arbitrarily ...

5

The parameter $a$ is independent of temperature because we defined it this way. There is no other reason. Indeed, van der Waals equation is very much not a law of nature. It is just an empirical approximation. There are other approximations, maybe a dozen or more. You may create your own, where $a$ would decrease with temperature. Wait, it's been done ...

5

The 2/3 factor is from the math of the kinetic theory of gases. Think of a cube with six faces (i.e. three orthogonal directions), the particles have equal probability of hitting a face in the x, y or z direction. The pressure is exerted equally among the three orthogonal directions, hence a factor of 1/3. Wikipedia also shows the steps in deriving the ...

4

The virial equation of state reads: $$Z = 1 + \left(b - \frac{a}{RT}\right)\frac{1}{V_\mathrm{m}} + \left(\frac{b}{V_\mathrm{m}}\right)^2 \cdots$$ Going by this equation, it seems like increasing the value of $b$ will increase $Z$, which means repulsive forces will increase, making the gas liquefaction difficult. Also for $a = 0$ and $b = 0$, the gas ...

4

The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas. As written in Atkins Physical Chemistry:

4

You need to consider the $b$ term, as well as $a$. Rearranging the van der Waals equation to solve for $p$ gives $$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$ You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result. But a positive $b$ will act to increase the pressure compared to the ideal ...

4

What is confusing you is the thermodynamic concept of equilibrium. A system at equilibrium has no external source of energy flowing onto it and is perfectly mixed internally. These are not common circumstances in the real world. In a system that is in equilibrium, a solid and its corresponding liquid can exist at one temperature. Any extra energy added will,...

4

For a system in equilibrium: Think of it as a closed system, with a certain amount of energy shared between the solid and liquid components (discounting the container). At any instant, some of the faster-moving molecules of the liquid could hit the ice, melting a bit, and thereby losing some of their own energy and freezing. It takes ~80 cal/g to change ...

4

$\ce{H2+ (aq)}$ ions don’t exist. Furthermore, even if they did, they would not make a neutral compound $\ce{H2SO4}$ with the sulfate ion, since the sulfate ion has a $2-$ charge. So you must split $\ce{H2SO4(aq)}$ into $\ce{2 H+}$ and $\ce{SO4^2-}$ ions.

3

Yes, this equation of state is original. While I cannot absolutely guarantee that it has never been used or proposed before, I can say that I have not encountered it in any “classical” model — and I teach thermodynamics and statistical physics, so I have seen many textbooks and classical exercices :) Extending beyond the question of originality, I read your ...

3

According to this source: The popular EoSs such as Soave-Redlich-Kwong (SRK) [1] and Peng-Robinson (PR) [2] predict liquid density with an average absolute error of about 8%, much higher than several good density correlations. This large magnitude of error is not acceptable by industry; therefore, they are not used for this purpose. Which is what I ...

3

There are different factors contributing to the pVT state, and some of them have (different!) temperature dependencies and are interacting. volume of particles electrostatic repulsion of electron shells London dispersion dipolar interaction energy distribution of particles kinetic energy rotational excitation (quantised, starts at moderate temperatures) ...

3

Using the corresponding states approach (normalizing to the critical pressure and critical temperature) usually gives pretty accurate results for design purposes.

3

Isn't that a bit different? At 293 K, you need about 56 bar to liquefy carbon dioxide. If you then release the pressure, a part of the liquid evaporates again while cooling down the rest until it solidifies at around -80 °C.

3

The expression you've written can be obtained from the equipartition theorem $$U = n_\mathrm{d.f.} \times \frac{k_\mathrm B T}2$$ stating that the energy is evenly distributed over all available degrees of freedom $n_\mathrm{d.f.}$, where each degree of freedom contributes an amount of energy equal to $k_\mathrm B T/2$. For a monoatomic gas of $N$ particles, ...

3

Collision is predominantly a bimolecular act. The probability of three gas molecules simultaneously colliding is incredibly low, therefore it quite accurate to say that the excluded molar volume $b$ is overcounted by a factor of 2.

3

I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated. Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP

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