Question

For a real gas $\text{ }\left( \text{mol}\text{.mass = 60} \right)$ if density at a critical point $\text{ 0}\text{.80 g/c}{{\text{m}}^{\text{3 }}}$ and it's $\text{ }{{\text{T}}_{\text{C}}}\text{ = }\dfrac{4\times {{10}^{5}}}{821}\text{K }$ then van der Waals constant a (in $\text{ atm }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{-2\text{ }}}$ ) is
A) $\text{ 0}.3375\text{ }$
B) $\text{ }3.375\text{ }$
C) $\text{ 1}\text{.68 }$
D) $\text{ 0}\text{.025 }$

Answer 1

The critical temperature of a gas is a temperature at and above which the vapour of the substance cannot be liquefied irrespective of the pressure applied to the substance. The critical temperature $\text{ }{{\text{T}}_{\text{C }}}$ is related to the van der Waals constant ‘a’ and ‘b’ is stated as follows,
$\text{ }{{\text{T}}_{\text{C }}}\text{= }\dfrac{\text{8a}}{\text{27Rb}}\text{ }$
Where R is the gas constant.

Complete Solution :
The gases can be liquefied by compressing the gas at a suitable temperature. The liquefaction of gases becomes difficult as the temperature of the gases increases. The increase in temperature increases the kinetic energy of gas particles.
- The critical temperature of a gas is a temperature at and above which the vapour of the substance cannot be liquefied irrespective of the pressure applied on the substance.
- The pressure which is needed to liquefy the gas at critical temperature is known as critical pressure and volume as critical volume.
The critical temperature $\text{ }{{\text{T}}_{\text{C }}}$ is related to the van der Waals constant ‘a’ and ‘b’ is stated as follows:
$\text{ }{{\text{T}}_{\text{C }}}\text{= }\dfrac{\text{8a}}{\text{27Rb}}\text{ }$ (1)
We are interested to determine the value of van der Waals constant ‘a’. Let's first determine the value of van der Waals constant ‘b’ .the critical volume $\text{ }{{\text{V}}_{\text{C}}}$ is equal to the three times of the van der Waals constant ‘b’. That is,
$\text{ }{{\text{V}}_{\text{C}}}\text{ = 3 b }$
The critical volume of a gas is calculated as follows,
$\text{ }{{\text{V}}_{\text{C}}}=\dfrac{\text{MW}}{\text{density}}=\dfrac{60}{0.80}=\text{ 75 c}{{\text{m}}^{\text{3}}}\text{mo}{{\text{l}}^{-1}}\text{ }$
Let’s determine the value of van der Waals constant ‘b’. We have,
$\begin{aligned} & \text{ }{{\text{V}}_{\text{C}}}\text{ = 3 b } \\\ & \Rightarrow \text{b}=\dfrac{{{\text{V}}_{\text{C}}}}{3}=\dfrac{75}{3}\text{ = 25 c}{{\text{m}}^{\text{3}}}\text{ mo}{{\text{l}}^{-1}}=0.0\text{25 L mo}{{\text{l}}^{-1}} \\\ \end{aligned}$
- Now let’s substitute the values in equation (1) we have,
$\begin{aligned} & \text{ }{{\text{T}}_{\text{C }}}\text{= }\dfrac{\text{8a}}{\text{27Rb}}\text{ } \\\ & \Rightarrow \dfrac{4\times {{10}^{5}}}{821}=\dfrac{8\times a}{27\times 0.0821\times 0.025} \\\ & \Rightarrow a=\dfrac{27\times 0.0821\times 0.025\times 4\times {{10}^{5}}}{821\times 8}=\dfrac{22167}{6568}=3.375\text{ atm }{{\text{L}}^{\text{2}}}\text{ mo}{{\text{l}}^{-2\text{ }}} \\\ \end{aligned}$
Therefore, the value of van der Waals constant ‘a’ is equal to $\text{ }3.375\text{ }$ .
So, the correct answer is “Option B”.

Note: Note that, by knowing the value of the critical constant of gas like critical pressure, temperature, and volume it is possible to calculate the values of van der Waals constant and vice versa. Here we have determined the critical volume $\text{ }{{\text{V}}_{\text{C}}}$ of gas by using the general formula of density. According to which density of a substance is mass per unit volume.