Equation of states for a gas that forms dimers

[Buffu]

Homework Statement

Show that to a first approximation the equation of state of a gas that dimerizes to a small extent is given by,

$\dfrac{PV}{RT} = 1 - \dfrac{K_c}{V}$

Where $K_c$ is equilibrium constant for $A + A \iff A_2$

Homework Equations

The Attempt at a Solution

Using virial expansion, I get

$\dfrac{PV}{RT} = 1 + \dfrac{B}{V} + \cdots$

Neglecting higher order terms,

I need to prove $B = -\dfrac{[A]^2}{[A_2]} = - K_c$,

I know how to compute $B$ from already known equation of states but I don't know any relation between $B$ and $K_c$.

Any hints please :).

 

[Chestermiller]

Homework Statement

Show that to a first approximation the equation of state of a gas that dimerizes to a small extent is given by,

$\dfrac{PV}{RT} = 1 - \dfrac{K_c}{V}$

Where $K_c$ is equilibrium constant for $A + A \iff A_2$

Homework Equations

The Attempt at a Solution

Using virial expansion, I get

$\dfrac{PV}{RT} = 1 + \dfrac{B}{V} + \cdots$

Neglecting higher order terms,

I need to prove $B = -\dfrac{[A]^2}{[A_2]} = - K_c$,

I know how to compute $B$ from already known equation of states but I don't know any relation between $B$ and $K_c$.

Any hints please :).

Treat the gas as ideal, but with the number of moles changing.

 

[Buffu]

Treat the gas as ideal, but with the number of moles changing.

Let the amount of gas in moles be $x$ and intial amount be $n$ moles.

Then $[A] = x/V$ and $[A_2] = (n -x)/V$

So, $K_c = \dfrac{x^2}{V(n -x)}$ --- (1)

also $\dfrac{PV}{RT} = x$ --- (2)

Should I solve for $x$ in $(1)$ and susbtitute it in $(2)$ ?

This does not feel correct.

 

[Chestermiller]

What is Kc supposed to be? Is it supposed to be based on concentrations or partial pressures? The Kc as you have written it is based on $$A_2---> 2A_1$$. Is that what it is supposed to be?

 

[Buffu]

What is Kc supposed to be? Is it supposed to be based on concentrations or partial pressures? The Kc as you have written it is based on $$A_2---> 2A_1$$. Is that what it is supposed to be?

Concentrations.

 

[Chestermiller]

OK. Here goes. At very large specific volumes, the contents of the tank will be all A, but, as the specific volume is decreased, some A₂ will be formed at the expense of A. Let $V_T$ be the current volume of the tank, and let $n_0$ be the number of moles of A that would be present in the tank if $V_T$ became very large. At volume $V_T$, suppose that there are now x moles of $A_2$. Then the moles of A will be $n_0-2x$, and the total number of moles will be $n=n_0-x$. So, the concentrations of A and $A_2$ when the volume is $V_T$ will be:

$$[A]=\frac{(n_0-2x)}{V_T}$$
$$[A_2]=\frac{x}{V_T}$$

These concentrations are related by the equilibrium constant for the reaction:
$$\frac{[A_2]}{[A]^2}=\frac{xV_T}{(n_0-2x)^2}=K_C$$If we neglect 2x in comparison to $n_0$, we can solve this equation for x, and obtain:
$$x=n_0^2\frac{K_C}{V_T}$$So the total number of moles at volume $V_T$ is given by:
$$n=n_0-n_0^2\frac{K_C}{V_T}=n_0\left(1-\frac{K_C}{(V_T/n_0)}\right)=n_0\left(1-\frac{K_C}{V}\right)$$where V is the apparent specific volume $V_T/n_0$