Question

The value of critical temperature in terms of Van der Waal’s constant a and b is given by:
${\text{A}}{\text{. }}{T_c} = \dfrac{a}{{2Rb}} \\\ B.{\text{ }}{T_c} = \dfrac{a}{{27Rb}} \\\ C.{\text{ }}{T_c} = \dfrac{{8a}}{{27Rb}} \\\ D.{\text{ }}{T_c} = \dfrac{{17a}}{{8Rb}} \\\$

Answer 1

The critical temperature is said to be a substance is the temperature at and above which vapor the substance cannot be liquefier in which no matter how much pressure is applied. Every substance has a critical temperature.

Complete step-by-step answer :
A Vander Waal equation is given by:
$\left( {P + \dfrac{a}{{{V^2}}}} \right)(V - b) = RT$
Where a and b is constant
By solving above equation we get,
$P = \dfrac{{RT}}{{V - b}} - \dfrac{a}{{{V^2}}}$
Taking the derivatives of the P with respect to volume we get,
$\dfrac{{\partial P}}{{\partial V}} = 0 \\\ \dfrac{{{\partial ^2}P}}{{\partial {V^2}}} = 0 \\\ \\\$
Now P, becomes;

$$\dfrac{{\partial P}}{{\partial V}} = - \dfrac{{RT}}{{V - b}} - \dfrac{a}{{{V^2}}} = 0 \\\ \Rightarrow \dfrac{{2a}}{{{V^3}}} = - \dfrac{{RT}}{{{{(V - b)}^2}}}...............................(1) \\\ \Rightarrow \dfrac{a}{{{V^4}}} = - \dfrac{{RT}}{{2V{{(V - b)}^2}}}..........................(2) \\\$$

Taking the double derivative now, we get;
$\dfrac{{{\partial ^2}P}}{{\partial {V^2}}} = \dfrac{{2RT}}{{{{(V - b)}^3}}} - \dfrac{{6a}}{{{V^4}}} = 0 \\\ or \\\ \dfrac{{RT}}{{{{(V - b)}^3}}} = \dfrac{{3a}}{{{V^4}}} \\\$
Put the equation (2) in above equation;
$\dfrac{{RT}}{{{{(V - b)}^3}}} = \dfrac{{3RT}}{{2V{{(V - b)}^3}}} \\\$
Rearranging,
$3V - 3b = 2V \\\ {V_c} = 3b \\\ {V_c} = is{\text{ critical velocity}} \\\$
Now use this value in equation(1)
$\dfrac{{RT}}{{4{b^2}}} = \dfrac{{2a}}{{27{b^3}}} \\\ {T_c} = \dfrac{{8a}}{{27RT}} \\\$

Where the ${T_c}$ is known as critical temperature.
Van der Waal is said to be the relatively weak attractive forces that act on neutral atoms and the molecules that arise because of the electric polarization induced in each of the particles by the presences of other particles. There are three types of Van der Waal; dispersion (weak), dipole-dipole (medium) and hydrogen (strong).
The importance of critical temperature of a gas is a measure of the strength of the intermolecular forces of attraction. General example for the critical temperature is liquid- vapor critical point and the end point of the pressure- temperature curve which indicates the condition under which a liquid and its vapor can coexist.
Therefore option (C) is the right answer.

Note :As we studied that the gases can be converted to two liquids by compressing the gases at a suitable temperature in liquefy gases become more difficult as the temperature increases because the kinetic energy of the particles which make up the gases also increases.