Question

The critical volume of a gas obeying van der Waals equation is:
A. $\dfrac{8a}{27{{R}_{b}}}$
B. $\dfrac{a}{27{{b}^{2}}R}$
C. 3b
D. $\dfrac{q}{27Rb}$

Answer 1

The Vander Waals equation for a real gas is as follows.
$(P+\dfrac{a}{{{V}^{2}}})(V-b)=RT$
Here P = Pressure of the gas
V =Volume of the gas
R = Gas constant
T = Temperature of the gas
a = Attraction between the gas particles
b = the volume excluded

Complete answer:
- In the question it is asked to find about the critical volume of the gas obeying Vander Waals equation among the given options.
- By using the Vander Waals gas equation for real gas, we can find the critical volume of the gas which obeys Vander Waals gas equation.
- The Vander Waals gas equation is as follows.
$(P+\dfrac{a}{{{V}^{2}}})(V-b)=RT$
Here P = Pressure of the gas
V =Volume of the gas
R = Gas constant
T = Temperature of the gas
a = Attraction between the gas particles
b = the volume excluded
- We can rearrange the above Vander Waals gas equation as follows.
$P=\dfrac{RT}{V-b}-\dfrac{a}{{{V}^{2}}}$
- We know that at critical temperature $\dfrac{\partial P}{\partial V}=0$
- Therefore,

& \Rightarrow \dfrac{-RT}{{{(V-b)}^{2}}}-\dfrac{2a}{{{V}^{3}}}=0 \\\ & \Rightarrow \dfrac{2a}{{{V}^{3}}}=\dfrac{RT}{{{(V-b)}^{2}}}--(1) \\\ \end{aligned}$$ \- And $$\dfrac{{{\partial }^{2}}P}{\partial {{V}^{2}}}=0$$ $$\Rightarrow \dfrac{6a}{{{V}^{4}}}=\dfrac{2RT}{{{(V-b)}^{3}}}--(2)$$ \- By dividing the equation (2) with (1) we will get the required equation and it is as follows. $$\begin{aligned} & \Rightarrow \dfrac{3}{V}=\dfrac{2}{V-b} \\\ & \Rightarrow V=3b \\\ \end{aligned}$$ \- Therefore, the critical volume of a gas obeying van der Waals equation is, V = 3b **So, the correct option is C.** **Note:** We must do partial differentiation of the Vander Waals equation for two times to get the critical volume of the gas which is obeying the Vander Waals equation. At time of partial differentiation, we must consider, $\dfrac{\partial P}{\partial V}=0$.