Question

The ratio $\dfrac{{\text{a}}}{{\text{b}}}$ (the terms used in van der Waals equation) has the unit:
A) ${\text{atm}}\,\,{\text{liter}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
B) ${\text{atm}}\,\,{\text{d}}{{\text{m}}^3}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
C) ${\text{dyne}}\,{\text{cm}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
D) All of the above

Answer 1

The ideal gas equation is obtained from the van der Waals equation where constants a and b are zero in the ideal gas equation. Here, the constants a and b are also called van der Waals constants. These constants are correction terms a gives correction to the intermolecular forces while b gives the correction to the molecular size. When these constants are zero then the ideal gas equation is ${\text{pV = nRT}}$ obtained.

Formula used: The van der Waals’s equation is as follows:
$\left[ {{\text{P}}\, + \dfrac{{{\text{a}}{{\text{n}}^2}}}{{{{\text{V}}^{\text{2}}}}}} \right]\left[ {{\text{V}} - {\text{nb}}} \right]{\text{ = nRT}}$ (i)
Here, the pressure is P, the volume is V, the number of molecules is n, the gas constant is R, the temperature is T, and van der Waals constants are a and b.

Complete step-by-step answer:
Now, use the van der equation given above,
$\left[ {{\text{P}}\, + \dfrac{{{\text{a}}{{\text{n}}^2}}}{{{{\text{V}}^{\text{2}}}}}} \right]\left[ {{\text{V}} - {\text{nb}}} \right]{\text{ = nRT}}$
Now, we have to rearrange the equation for P as follows:
${\text{P}} = \dfrac{{{\text{nRT}}}}{{{\text{V}} - {\text{nb}}}} - \dfrac{{{{\text{n}}^{\text{2}}}{\text{a}}}}{{{{\text{V}}^{\text{2}}}}}$ (ii)

As we know that if two terms have the same units then and then only it gets added.
Here, the term ${\text{V}} - {\text{nb}}$ indicates ${\text{nb}}$ has the same unit as that of volume. Now, determine the unit of the b as follows:
${\text{unit}}\,{\text{of}}\,{\text{b = }}\dfrac{{{\text{unit of V}}}}{{{\text{unit}}\,{\text{of n}}}}$
${\text{unit}}\,{\text{of}}\,{\text{b = }}\dfrac{{{\text{liter}}}}{{{\text{mole}}}}$……(iii)
Now, determine the unit of the constant a as follows:
From the above equation, we can say that the units of P and $\dfrac{{{{\text{n}}^{\text{2}}}{\text{a}}}}{{{{\text{V}}^{\text{2}}}}}$ are the same. Now, determine the unit of a as follows:

${\text{unit}}\,{\text{of}}\,{\text{P}}\,{\text{ = }}\,{\text{unit}}\,{\text{of }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{a}}}}{{{{\text{V}}^{\text{2}}}}}$
${\text{atm}}\,{\text{ = }}\,\dfrac{{{\text{mol}}{{\text{e}}^{\text{2}}} \times {\text{unit}}\,{\text{of}}\,{\text{a}}}}{{{\text{lite}}{{\text{r}}^{\text{2}}}}}$
${\text{unit}}\,{\text{of}}\,{\text{a}} = \dfrac{{{\text{atm}} \times \,{\text{lite}}{{\text{r}}^{\text{2}}}}}{{{\text{mol}}{{\text{e}}^{\text{2}}}}}$ (iv)
Now, to determine the unit of $\dfrac{{\text{a}}}{{\text{b}}}$ taking the ratio of equation number (iv) to (iii),
$\dfrac{{{\text{unit}}\,{\text{of}}\,{\text{a}}}}{{{\text{unit}}\,{\text{of}}\,{\text{b}}}} = \dfrac{{{\text{atm}} \times \,{\text{lite}}{{\text{r}}^{\text{2}}} \times {\text{mole}}}}{{{\text{mol}}{{\text{e}}^{\text{2}}} \times {\text{liter}}}}$
$\dfrac{{{\text{unit}}\,{\text{of}}\,{\text{a}}}}{{{\text{unit}}\,{\text{of}}\,{\text{b}}}} = \dfrac{{{\text{atm}} \times \,{\text{liter}}}}{{{\text{mole}}}}$
Thus, the ratio obtained is ${\text{atm}}\,\,{\text{liter}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$.
Therefore, option (A) is correct.

Here, option (B) given is ${\text{atm}}\,\,{\text{d}}{{\text{m}}^3}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$in this volume is given in decimetre.
${\text{1}}\,{\text{liter}}\,{\text{ = }}\,{\text{1d}}{{\text{m}}^3}$
${\text{atm}}\,\,{\text{d}}{{\text{m}}^3}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}} = {\text{atm}}\,\,{\text{liter}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
Therefore, option (B) is also the correct answer to the question.
Here, option(C) is ${\text{dyne}}\,{\text{cm}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ it is also a correct answer.
Because ${\text{atm}}\,\,{\text{liter}}\,\,{\text{ = }}\,{\text{conversion factor}} \times {\text{dyne}}\,{\text{cm}}$
Therefore, ${\text{atm}}\,\,{\text{d}}{{\text{m}}^3}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}} = {\text{atm}}\,\,{\text{liter}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}} = {\text{dyne}}\,{\text{cm}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

Therefore, option (D) all of these the above is the correct answer for the given question.

Note: In any equation, if two terms are added or subtracted then their units must be equal otherwise there is no addition of the terms. In the case of van der Waals equation to determine the units of the constants a and b rearrange the equation for pressure and then compare the units.