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The unit of the van der Waals constant '$a$' in the equation $\left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$ is:
In the van der Waals equation, the term $\dfrac{an^2}{V^2}$ must have units of pressure.
So: $a = \dfrac{P \cdot V^2}{n^2}$
Units of $a = \dfrac{\text{atm} \cdot (\text{dm}^3)^2}{(\text{mol})^2} = \mathbf{\text{atm dm}^6\ \text{mol}^{-2}}$
The constant $b$ has units of dm$^3$ mol$^{-1}$ (it represents excluded molar volume). 'a' corrects for intermolecular attractions.
Suppose $x, y, z$ are in arithmetic progression (A.P.) and $\tan^{-1}x,\ \tan^{-1}y,\ \tan^{-1}z$ are also in A.P. Then:
If $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are in A.P., then $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z$.
Also $x,y,z$ in A.P. means $2y=x+z$.
From $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z$: using the addition formula, $\tan(2\tan^{-1}y)=\tan(\tan^{-1}x+\tan^{-1}z)$, i.e., $\dfrac{2y}{1-y^2}=\dfrac{x+z}{1-xz}$.
Substituting $x+z=2y$: $\dfrac{2y}{1-y^2}=\dfrac{2y}{1-xz}$. So $1-y^2=1-xz$, giving $y^2=xz$.
Combined with $x+z=2y$ (A.P.) and $y^2=xz$ (G.P.) $\Rightarrow x=y=z$.
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