Wednesday, 19 October 2016

physical chemistry - Does the van der Waals equation remain valid when repulsive intermolecular forces dominate?


The van der Waals equation for a real gas is:


$$RT =\left(p+\frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m}-b)$$


We have understood this formula by saying that $a$ is the term which is for force of attraction between the gas molecules and thus the pressure of a real gas will be less than that of an ideal gas.


Therefore some factor must be added to the pressure of a real gas to make it equal to ideal gas pressure so that it can be used in $p_\mathrm{ideal}V_{\mathrm{ideal}}= nRT$.


Also, if we see the compressibility factor graph , suppose for $\ce{N2}$:


Graph of Z vs P for N2


(Source: TutorVista)


If $Z < 1$, then attractive forces are dominating. We see that after a particular temperature, the value of $Z$ becomes greater than $1$, implying that the repulsion forces are dominating in $\ce{N2}$.


Now, since repulsive force dominates, the pressure of the real gas will increase and be greater than that in ideal case.



To find the $p_{\mathrm{ideal}}$ now, we will need to deduct some value from the $p$ of real gas, which means $a$ should be negative. However, this is not so as $a$ can never be negative.


No value can ever be deducted from $p$ in the van der Waals equation.


Then how does the van der Waals equation still work and give the correct data?




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