14

Yes, they are certainly correlated as shown in the following plot based on data from the Wikipedia Van der Waals constants data page The correlation should not be entirely surprising: the covolume b is a measure of the size of the gas molecules, whereas a is a measure of the strength of intermolecular interactions which is expected to depend through ...


7

The van der waals constant can be directly obtained from the critical temperature and pressure. We have, ${a = \frac{27 R^2 T_c^2} {64 P_c}}$ ${b = \frac{R T_c }{ 8 P_c}}$ Since, $T_c$ and $P_c$ are constants, we can conclude that the van der waals constants are really "constants."


7

A simple demonstration in electrostatics can convince that London dispersion forces are attractive. Has your physics teacher shown "charging by induction"? Charging by induction video Now replace the rods by molecules. Note the rods always attract. If one molecule develops a dipole or negative charge (on one side), it will induce an oppositive ...


6

I would add two points to MFarooqs' answer. First, I would emphasize that the basis for attractive van der Waals interactions is the polarizability of molecules. Polarizability is the property of having flexible charge distributions which may be distorted by interacting with charges outside of a molecule, leading to a more stable electronic state. ...


6

As Ivan writes $a$ and $b$ are independent, but it does not mean that they are not correlated; however, so are a child's shoe size and reading ability which clearly does not mean that buying a child larger shoes will increase its reading ability. A correlation does not imply cause and effect. Plots of the compression factor with reduced pressure show that ...


6

Just for curiosity, I also plotted a vs b using the same date set used by Buck Thorn. What I want to know is what is the correlation shown in Buck Thorn's plot and what are the two outliers clearly in the plot: As I marked in my plot, they are mercury and idobenzene. As Karl pointed out in one of his comments, mercury is the biggest outlier. Other than ...


6

$$Z = \frac{V^{\mathrm{r}}_{\mathrm{m}}}{V^{\circ}_{\mathrm{m}}}\tag{1}\label{1}$$ $V^{\mathrm{r}}_{\mathrm{m}} = \text{Volume of 1 mol real gas.}$ $V^{\circ}_{\mathrm{m}} = \text{Volume of 1 mol perfect gas.}$ The van der Waals equation is $$ \left( P^{\text{r}} + \frac{an^2}{(V^{\text{r}})^2}\right) (V^{\text{r}} - nb ) = nRT \tag2\label{2}$$ Now, ...


6

Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state. The perfect gas equation, $ PV = nRT $ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces ...


5

The constants a and b in the Van der Waals equation are supposed to be independent of temperature. But it is important to remember that, even though the Van der Waals equation does a better job of approximating the behavior of real gases than the ideal gas law (over a larger range of parameter values), it too is just approximation that applies only over a ...


5

At the critical point, the first and second derivatives of pressure with respect to volume are zero. This provides two equations to solve for the two unknowns (a and b) in terms of the critical temperature and critical volume (or pressure). See Critical Constants of the van der Waals Gas


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 ...


3

There are 2 deviation terms of the opposite sign. The negative deviation term relates to the attractive forces between molecules, decreasing pressure,and is directly related to condensation ability of real gases. The positive deviation term relates to non-negligible own volume of molecules, compared to the total gas volume, leading to higher than ideal ...


3

Intermolecular attraction can reduce the volume, in the same way that increasing the pressure can lower the volume (Boyle's law). One way to see the effect is to take the derivative of the volume, as given by the van der Waals (vdW) equation, with respect to the attractive parameter $a.$ The vdW equation is $$\left(p+\frac{a}{\bar{V}^2}\right)(\bar{V}-b)=RT\...


2

When you see the ideal gas equation, derieved from the kinetic theorey of gases, there are many assumptions taken for the ideal gases, a few like their Potential energy is 0, the molecules are spherical, and so on. What different real gas equations attempt to do is to reduce the number of assumptions taken by providing "corrections". Van der Waal attempted ...


2

Your analysis is correct in terms of number density. But let's see how it plays out in terms of molar density. Let n be the number of moles and A be Avagadro's number. Then N=nA. If we substitute this into your first equation, then I get $$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where ...


2

There are dozens of "real gas equations" which try to rationalize one factor or another. Wall effects are never one of the problems considered. Such real gas equations are for the "bulk" gas not what is happening at the surface. Wall effects are generally ignored entirely... (1) In general the notion is that the wall collisions are "on average" an ...


2

From the given data, the molar volume $V_\mathrm m$ is $$V_\mathrm m=\frac{150\ \mathrm{cm^3}}{3\ \mathrm{mol}}=50\ \mathrm{cm^3/mol}=0.05\ \mathrm{L/mol}$$ In terms of molar volume, the van der Waals equation reads: $$\left(p+\frac{a}{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$ So, solving for the pressure gives: $$p=\frac{RT}{V_\mathrm m-b}-\frac{a}{V_\...


1

The more general equation you are looking for is called the thermodynamic equation of state. You can start from the differential form of U according to the 1st + 2nd laws of thermodynamics for pV work only: $$dU = TdS-pdV$$ Taking the derivative wrt V at constant T $$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial S}{\partial V}\...


1

The surface tension of water is $\pu{72 dyne/cm}$; it is sometimes compared to mercury, which has a surface tension of $\pu{486 dyne/cm}$. The higher surface tension of the mercury is because the atoms of mercury bond much more tightly. A molecule at the surface of a liquid experiences net inward cohesive forces. Usually, we try to reduce those forces in ...


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