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<math>
<math>
\dfrac{\partial^2 P}{\partial V^2} \bigg|_{T=T_c,V=V_c} = 0
\dfrac{\partial^2 P}{\partial V^2} \bigg|_{T=T_c,V=V_c} = 0
</math>
=Question 1=
Show that the critical point <math>(P_c, V_c, T_c)</math> is given by:
<math>
P_c = \dfrac{a}{27b^2}
</math>
<math>
V_c = 3b
</math>
<math>
T_c = \dfrac{8a}{27bk}
</math>
</math>

Revision as of 03:14, 22 May 2016

Background

Critical points of gas, supercritical behavior

The Van Der Waal equation for a gas accounts for non-ideal behavior caused by strong intermolecular forces of attraction or repulsion:

$ (P + \dfrac{a}{V^2})(V-b) = kT $

where P is the pressure, V is the molar volume (volume of a certain number of moles), a and b are constants that depend on the molecules, k is the Boltzmann constant, and T is temperature.

Now the critical points can be found: https://www.youtube.com/watch?v=VjVQxzxxLVw

Critical point is the saddle point of the above equation, and is defined as the point where:

$ \dfrac{\partial P}{\partial V} \bigg|_{T=T_c,V=V_c} = 0 $

and

$ \dfrac{\partial^2 P}{\partial V^2} \bigg|_{T=T_c,V=V_c} = 0 $

Question 1

Show that the critical point $ (P_c, V_c, T_c) $ is given by:

$ P_c = \dfrac{a}{27b^2} $

$ V_c = 3b $

$ T_c = \dfrac{8a}{27bk} $