What Is The Method To Combine Two Mathematical Conditions To Get A Condensed Equation/function

How can we calculate critical temperature, volume and pressure in terms of a and b?

This can be done from the Van der Waals Equation of State for real gases. Consider that equation, for one mole of the gas in question:[math](P+\frac{a}{V^2})(V-b)=RT[/math]A bit of rearrangement gives us-[math]P=\frac{RT}{V-b}-\frac{a}{V^2}[/math]Now if a graph of P as a function of volume were to be plotted, it would be something like this-As you can see, the critical temperature [math](T_c)[/math] also happens to be a point of inflection for the PV diagram. One concludes, at this temperature-[math]\frac{\partial P}{\partial V} = 0[/math][math]\frac{\partial^2 P}{\partial V^2}=0[/math]Where P is a function of V as per the Van der Waals' Equation. Making use of that, we get:[math]\frac{\partial P}{\partial V} = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3}=0[/math][math]\Rightarrow \frac{2a}{V^3} = \frac{RT}{(V-b)^2} . . . . . . (A_1)[/math][math]\Rightarrow \frac{a}{V^4} = \frac{RT}{2V(V-b)^2} . . . . . . (A_2)[/math]Also, by differentiating the first partial derivative again, with respect to V, we get:[math]\frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V-b)^3} - \frac{6a}{V^4} = 0[/math]Dividing both sides by 2, and taking one term to the left hand side, we get:[math]\frac{RT}{(V-b)^3} = \frac{3a}{V^4}[/math]Plugging in the value that we just obtained from Equation [math]A_2[/math]-[math]\frac{RT}{(V-b)^3} = \frac{3RT}{2V(V-b)^2}[/math]Now, if we divide both sides by a factor of [math]\frac{RT}{(V-b)^2}[/math], and just rearrange the terms, we see that-[math]3V - 3b = 2V[/math][math]\Rightarrow V_c = 3b, [/math]which is the critical volume.Plugging this value into equation [math]A_1[/math]-[math]\frac{RT}{4b^2} = \frac{2a}{27b^3}[/math][math]\Rightarrow T_c = \frac{8a}{27Rb},[/math]which is the critical temperature.To calculate the critical pressure, return to the original equation-[math]P=\frac{RT}{V-b}-\frac{a}{V^2}[/math]Plug in the values of [math]T_c[/math] and [math]V_c[/math] we just calculated, and you'll see that-[math]P_c =\frac{a}{27b^2}[/math]Image Source: Bright Hub Engineering.