Question

For a real gas(MW=60) if density at critical point is 0.80 $\dfrac{g}{{c{m^3}}}$ and its \[{T_c} = \dfrac{{4 \times {{10}^5}}}{{821}}K\] , than Vander Waals constant a (in atm \[{L^2}mo{l^{ - 2}}\] ) is:
A.0.3375
B.3.375
C.1.68
D.0.025

Answer 1

Hint: We can calculate the volume of a gas using the basic formula of density which is mass by volume. This will be the critical volume at critical temperature. It is related to the b constant of Vander Waals equation. With the value of ‘b’ then known to us, we can determine the value of ‘a’ constant using the value of critical temperature given in the question.
Formula used:
 \[{V_c} = \dfrac{{molar\,\,mass}}{{density}}\] and \[b = \dfrac{{{V_c}}}{3}\]

Complete step by step answer:
Vander Waals equation can be recalled as a modified version of an ideal gas law which states that the gases consist of point masses that can undergo perfectly elastic collisions. But it failed to explain the behaviour of real gases. That’s why the Vander Waals equation was devised and helped us to define the physical state of a real gas.
We are familiar with the ideal gas equation which is \[PV = nRT\] . For real gas, the Vander Waals equation for one mole can be written as:
 \[\left( {P + \dfrac{a}{{{V^2}}}} \right)(V - b) = RT\]
Where, a and b are constants specific to each gas and are known as Vander Waals constants. They always have a positive value. The constant ‘a’ represents the magnitude of intermolecular forces of attraction and constant ‘b’ represents the effective size of the molecules.
We know that critical volume is the volume occupied by the fluid or substance in its critical state and critical temperature is the temperature of a substance beyond which it cannot be liquefied. We are familiar with the very basic formula of density which is mass by volume. Therefore, volume will be
 \[{V_c} = \dfrac{{molar\,\,mass}}{{density}}\]
Putting the values of molar mass as 60gms and density at critical point is 0.80 $\dfrac{g}{{c{m^3}}}$ , we get
 \[{V_c} = \dfrac{{60}}{{0.80}} = 75c{m^3}mo{l^{ - 1}}\]
From the real gas equation, the constant b is related to critical volume such that the critical volume for 1 molecule of gas is equal to 3b.
 \[b = \dfrac{{{V_c}}}{3} = \dfrac{{75}}{3} = 25c{m^3}mo{l^{ - 1}}or\,0.025Lmo{l^{ - 1}}\]
 \[\therefore {T_c} = \dfrac{{8a}}{{27Rb}}\] is the critical temperature when critical volume is already known to us. Let us put the values of R and b in it and we will get the value of constant a.
 \[
  \dfrac{{4 \times {{10}^5}}}{{821}} = \dfrac{{8a}}{{27 \times 0.0821 \times 0.025}} \\
  a = 3.375 \\
 \]

Hence the correct option is (B).

Note:
If constants ‘a’ and ‘b’ of Vander Waals equation are small, then the terms \[\dfrac{a}{{{V^2}}}\] and b are neglected as compared to P and V. This reduces the equation back to the ideal gas equation, \[PV = nRT\] . The units of constants ‘a’ and ‘b’ are different such that unit of a is \[li{t^2}mo{l^{ - 2}}\] and unit of b is \[litremo{l^{ - 1}}\].