Question

The relationship between $Pc,\text{ }Vc$ and $Tc$ is:
A.${{P}_{c}}{{V}_{c}}=R{{T}_{c}}$
B.${{P}_{c}}{{V}_{c}}=3R{{T}_{c}}$
C.${{P}_{c}}{{V}_{c}}=\dfrac{3}{5}R{{T}_{c}}$
D.${{P}_{c}}{{V}_{c}}=\dfrac{3}{8}R{{T}_{c}}$

Answer 1

We know that recall the meaning of compressibility factor, critical temperature, pressure, and volume. Think about how all of these are related and expressed in terms of the Vander Waals constants.

Complete answer:
Here is the compressibility factor which shows the deviation of the behaviour of a real gas from the behaviour of an ideal gas. The Vander Waals constants ‘a’ and ‘b’ are used to calculate the value of at critical temperature, pressure, and volume.
${{T}_{c}}=~$ Temperature at and above which the vapour state cannot be further liquefied; ${{P}_{c}}=~$ Pressure of gas in its critical state; ${{V}_{c}}=~$ Volume of gas in its critical state.
Thus, we know that relation with the van der Waals constants -
${{T}_{c}}=\dfrac{8a}{27Rb~};a=\dfrac{27}{8}{{T}_{c}}Rb$ and ${{V}_{c}}=3b~$; $b=\dfrac{{{V}_{c}}}{3}$
Thus,${{P}_{c}}=\dfrac{a}{27{{b}^{2}}}...\left( 1 \right)$
Put the value of a and b in $equation\text{ }\left( 1 \right)$ we get;
$Pc=\dfrac{27{{T}_{c}}\times R\times b}{8\times 27\times {{b}^{2}}}$
On further solving we get;
${{P}_{c}}=\dfrac{27{{T}_{c}}R}{8\times 27\times b}=\dfrac{27{{T}_{c}}R}{8\times 27\times \dfrac{Vc}{3}}$
Further we get; ${{P}_{c}}{{V}_{c}}=\dfrac{3}{8}R{{T}_{c}}$

Therefore, the correct answer is option D.

Note:
Remember that the compressibility factor in the critical state of any real gas is known to be $~3/8$. The relation between the critical states and the Vander Waals constant is derived using Van Der Waals real gas equation.