# Dieterici's equation

1. Aug 2, 2007

### 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.

Last edited: Aug 2, 2007
2. Aug 3, 2007

### HallsofIvy

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

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

3. Sep 17, 2007

### eccles1214

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

4. Sep 17, 2007

### 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 . . .

5. Sep 27, 2007

### 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:
$$\frac{RT}{(v-b)^2}\exp^{\frac{-a}{vRT}} - \frac{a}{(v-b)v^2} \exp^{\frac{-a}{vRT}} \\ = \exp^{\frac{-a}{vRT}}(\frac{RT}{(v-b)^2} - \frac{a}{(v-b)v^2}) \\$$ $$\mbox{ For the critical volume } \ v_c \ \mbox{ and the critical temperature } \ T_c \\ \ \frac{{\partial p}}{{\partial v}}= \exp^{-2}( \frac{a}{4b(b)^2} - \frac{a}{b4b^2})=0$$