Question

The value of $\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}$
(A) $\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}=\dfrac{3}{8}$
(B) $\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}=\dfrac{5}{8}$
(C) $\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}=\dfrac{1}{2}$
(D) None of these

Answer 1

Hint: The value of $\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}$ depends upon the critical phenomenon of the gases.

Complete step by step solution:
We know that,
Van der Waals equation for one mole of real gas is,
$\left( p+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$
Now,
Let us see what does the critical phenomenon of gases mean-
The gas can be liquified by reducing the temperature and increasing the pressure. However, above a certain temperature, the gas cannot be converted to liquid by increasing pressure alone. Thus, temperature acts dominant to pressure changes taking place.
The temperature below the gas can be liquified is known as critical temperature of the gas and the pressure required at this condition to liquify the gas is known as critical pressure. The volume occupied by one mole of the gas under critical conditions is known as critical volume of the same.
Thus, using Van der Waals equation for one mole of a real gas,
$\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$
We can say that,
${{V}^{3}}-\left( b+\dfrac{RT}{P} \right){{V}^{2}}+\dfrac{a}{P}V-\dfrac{ab}{P}=0$
At critical conditions,
$P={{P}_{c}},T={{T}_{c}},V={{V}_{c}}$
Thus,
We can write;
$V-{{V}_{c}}=0$
Cubing on both sides we get,
${{V}^{3}}-3{{V}_{c}}{{V}^{2}}+3{{V}_{c}}^{2}V-{{V}_{c}}^{3}=0$
Thus, comparing above equations,
We get, ${{V}_{c}}=3b$
Similarly, doing above analysis with respect to pressure and temperature will result into;
${{P}_{c}}=\dfrac{a}{27{{b}^{2}}}$ and ${{T}_{c}}=\dfrac{8a}{27Rb}$
Hence, the Van der Waals constants for any gas can be calculated by using critical parameters of gas and vice versa.
Now,
$\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}$ = $\dfrac{\left( \dfrac{a}{27{{b}^{2}}}\times 3b \right)}{\left( R\times \dfrac{8a}{27Rb} \right)}$
$\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}$ = $\dfrac{3}{8}$

Therefore option (A) is correct.

Additional information:
This is also known as the critical compressibility factor of a gas.
i.e. ${{Z}_{c}}$ = $\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}$ = $\dfrac{3}{8}$ = 0.375
The equation of critical state can be-
${{P}_{c}}{{V}_{c}}=\dfrac{3}{8}R{{T}_{c}}$

Note: Van der Waals equation for real gas is derived from the Van der Waals equation of ideal gas. All gases are real in nature, so we should always use the real gas equation while calculating for the compressibility factor.