Question

For a real gas (mol. Mass 30) if density as critical point is $0.40{\text{ g}}/{\text{c}}{{\text{m}}^3}$ and ${{\text{T}}_{\text{C}}} = \dfrac{{2 \times {{10}^5}}}{{821}}{\text{K}}$ , then calculate van der waals constant a $\left( {{\text{in atm }}{{\text{L}}^2}{\text{ mo}}{{\text{l}}^{ - 2}}} \right)$

Answer 1

We must know the formula for critical temperature and pressure in terms of van der waal gas constant. Using the ideal gas equation we will get the value of pressure. Equating the equation we will find the value of one of the constants.
Formula used:
${{\text{T}}_{\text{C}}} = \dfrac{{8{\text{a}}}}{{27{\text{Rb}}}}$ here ${{\text{T}}_{\text{C}}}$ is critical temperature, a is pressure correction term, b is volume correction factor and R is universal gas constant.
${{\text{P}}_{\text{C}}} = \dfrac{{\text{a}}}{{27{{\text{b}}^2}}}$ here ${{\text{P}}_{\text{C}}}$ is critical pressure.
${\text{PV}} = {\text{nRT}}$ p is pressure, V is volume, n is number of moles, T is temperature

Complete step by step answer:
We have been given the value of critical temperature. We will now compare the given value with the formula of critical temperature. We will get:
$\dfrac{{8{\text{a}}}}{{27{\text{Rb}}}} = \dfrac{{2 \times {{10}^5}}}{{821}}{\text{K}}$
We will now rearrange the above equation in terms of constant a:
$\dfrac{{8 \times 821{\text{a}}}}{{27{\text{R}} \times 2 \times {{10}^5}}} = {\text{b}}$
Let us keep the above equation for further use. Now we know the formula for ideal gas equation:
${\text{PV}} = {\text{nRT}}$
Now the number of moles is the ratio of mass of the substance to its molar mass. We will represent mass by m and molar mass by M. So we will get the equation as by rearranging:
${\text{P}} = \dfrac{{\text{m}}}{{{\text{M}} \times {\text{V}}}}{\text{RT}}$
Now mass per unit volume is called density. So we will introduce density here by representing it with d.
${\text{P}} = \dfrac{{\text{d}}}{{\text{M}}}{\text{RT}}$
Now we will equate the above equation with the formula for critical pressure and we will get:
$\dfrac{{\text{d}}}{{\text{M}}}{\text{RT}} = \dfrac{{\text{a}}}{{27{{\text{b}}^2}}}$
Keeping the constant a on one side and substituting the value we will get:
${\text{a}} = {\left( {\dfrac{{27 \times 0.0821 \times 2 \times {{10}^5}}}{{8 \times 821}}} \right)^2} \times \dfrac{{30 \times 821}}{{400 \times 0.0821 \times 27 \times 2 \times {{10}^5}}}$
Rearranging it further we will get:
${\text{a}} = 0.633{\text{ atm }}{{\text{L}}^2}{\text{ mo}}{{\text{l}}^{ - 2}}$

Note:
According to the ideal gas equation there is no force of attraction between the gas molecules. But this is not observed in real life. Hence, the correction factor is introduced in the van der waal equation.