Dieterici's equation

[John O' Meara]

Dieterici's equation ( an alternative to Van der waal's equation) states that the pressure p, volume v and absolute temperature T of a mass of gas are connected by the equation
p=\frac{RT}{(v-b)}\exp{\frac{-a}{vRT}} \\ ,
where a, b and R are constants. Verify that both \frac{{\partial p}}{{\partial v}} = 0 \\
and \frac{{\partial^2 p}}{{\partial^2 v}} =0 \\, for the critical volume and temperature v_c and T_c respectively,where v_c = 2b and T_c = \frac{a}{4bR} \\ . What is the value of p_c the critical pressure in terms of a,b and e?
I have a question in solving this: namely is \frac{d\exp{\frac{-a}{vRT}}}{dv} = \frac{d \exp{\frac{-a}{vRT}}}{d v^{-1}} \frac{d v^{-1}}{dv} \\ Because I don't think so: could someone explain what the l.h.s. is equal to. Thanks for the help.

[HallsofIvy]

\frac{d exp(\frac{-\alpha}{vRT})}{dv}= exp(\frac{-\alpha}{vRT})\frac{d\frac{-\alpha}{vRT}}{dv}= exp(\frac{-\alpha}{vRT})\frac{\alpha}{v^2RT}

That's because
\frac{d \frac{A}{v}}{dv}= \frac{d Av^{-1}}{dv}= -Av^{-2}
for any constant A.

[eccles1214]

\frac{d exp(\frac{-\alpha}{vRT})}{dv}= exp(\frac{-\alpha}{vRT})\frac{d\frac{-\alpha}{vRT}}{dv}= exp(\frac{-\alpha}{vRT})\frac{\alpha}{v^2RT}


Ok, I finally got to this step. But how do I take the 2nd derivative of this last result? It's gnarly.

[eccles1214]

Ok,

I think I got the 2nd derivative, and then I set both 1st and 2nd derivative to zero.
Now I have 3 equations (original, 1st derivative, 2nd derivative), but how many unknowns? I know that V is an unknown, but isn't T also an unknown? I treated it as a constant.

Do I solve for V in the 1st derivative and plug it into the 2nd derivative?

This is all very confusing . . .

[John O' Meara]

\frac{{\partial p}}{{\partial v}} = \frac{{\partial}}{{\partial v}} (\frac{RT}{v-b}\exp^{\frac{-a}{vRT}}) \\
which = \exp^{\frac{-a}{vRT}} \frac{{\partial }}{{\partial v}}(\frac{RT}{v-b}) + \frac{RT}{v-b} \frac{{\partial }}{{ \partial v}}(\exp^{\frac{-a}{vRT}}) \\ .
Now use HallsofIvy's equation to evaluate the second term of the product rule expression to get the following:
\frac{RT}{(v-b)^2}\exp^{\frac{-a}{vRT}} - \frac{a}{(v-b)v^2} \exp^{\frac{-a}{vRT}} \\ = \exp^{\frac{-a}{vRT}}(\frac{RT}{(v-b)^2} - \frac{a}{(v-b)v^2}) \\ \mbox{ For the critical volume } \ v_c \ \mbox{ and the critical temperature } \ T_c \\ \ \frac{{\partial p}}{{\partial v}}= \exp^{-2}( \frac{a}{4b(b)^2} - \frac{a}{b4b^2})=0