Question

One way of writing the equation of state for real gases is $P\overline V = RT\left[ {1 + \dfrac{B}{{\overline V }} + \ldots } \right]$, where $B$ is a constant. An approximate expression for $B$ in terms of the Van der Waals’ constant $'a'$ and $'b'$ is $b - \dfrac{{na}}{{RT}}$. Then, the value of $n$ is _____.

Answer 1

The van der Waals equation is written as $[P + \dfrac{a}{{{{\overline V }^2}}}][\overline V - b] = RT$. Generate the value for $P$ from this equation. Multiply both the sides with $\overline V$ and make it simpler step by step.

Complete answer:
Van der Waals’ equation generalizes the ideal gas law, based on the fact that the real gases do not act ideally. According to the ideal gas law, The ideal gas law states that the volume $\left( V \right)$ occupied by $n$ moles of any gas having a pressure $\left( P \right)$ at temperature $\left( T \right)$ (in kelvin) is given by the following relationship:
$PV = nRT$ [Where, $R$ is the universal gas constant]
Now, as the real gas molecule takes up some amount of volume, the Van der Waals’ equation replaces the volume of the ideal gas with the molar volume of the gas $\overline V$, where, $b$ is the volume that is occupied by one mole of the molecules. So, the relation becomes
$P\left( {\overline V - b} \right) = RT$
The second modification was that Van der Waals provides the intermolecular interaction by adding the observed pressure $P$ in the equation with the term $\dfrac{a}{{{{\overline V }^2}}}$, where $a$ is the constant, whose value depends on the gas. Therefore, the van der Waals equation is written as:
$[P + \dfrac{a}{{{{\overline V }^2}}}][\overline V - b] = RT$
or, $P = \dfrac{{RT}}{{\overline V - b}} - \dfrac{a}{{{{\overline V }^2}}}$
Multiplying both the sides with $\overline V$, we get,
$P\overline V = \dfrac{{RT\overline V }}{{\overline V - b}} - \dfrac{{a \times \overline V }}{{{{\overline V }^2}}}$
or, $P\overline V = RT\left( {\dfrac{{\overline V }}{{\overline V - b}} - \dfrac{a}{{\overline V RT}}} \right)$ [Taking $RT$ common]
or, $P\overline V = RT\left[ {{{\left( {1 - \dfrac{b}{{\overline V }}} \right)}^{ - 1}} - \dfrac{a}{{\overline V RT}}} \right]$
Now, we know that, ${\left( {1 - \dfrac{b}{{\overline V }}} \right)^{ - 1}} = 1 + \left( {\dfrac{b}{{\overline V }}} \right) + {\left( {\dfrac{b}{{\overline V }}} \right)^2} + {\left( {\dfrac{b}{{\overline V }}} \right)^3} + .....$
So, $P\overline V = RT\left[ {1 + \dfrac{b}{{\overline V }} - \dfrac{a}{{\overline V RT}} + {{\left( {\dfrac{b}{{\overline V }}} \right)}^2} \ldots .} \right]$
or, $P\overline V = RT\left[ {1 + \left( {b - \dfrac{a}{{RT}}} \right).\dfrac{1}{{\overline V }} + {{\left( {\dfrac{b}{{\overline V }}} \right)}^2} + ....} \right]$
So, $B = b - \dfrac{a}{{RT}}$
So, the value of $n$ is $1$.

Note:
We should remember that when the molar volume $\overline V$ becomes large, $b$ becomes negligible with respect to the molar volume. Also, the term $\dfrac{a}{{{{\overline V }^2}}}$ becomes negligible with respect to the pressure $P$. Then, as a result, Van der Waals’ equation is reduced to the ideal gas law equation $P\overline V = RT$.