1
$\begingroup$

I'm studying thermodynamics and I found two forms for the virial expansion: $$pV~=~RT[1+\frac{A_2}{V}+\frac{A_3}{V^2}+\ldots] \tag{1}$$ and $$pV~=~RT[1+B_2p+B_3p^2+\ldots]\tag{2}$$ my problem is that I can not find the correct procedure to express the coefficients $B_k$ in terms of $A_k$. I just find the answer, that is $$B_2=\frac{A_2}{RT}, \qquad B_3=\frac{A_3-(A_2)^2}{(RT)^2}, \qquad \ldots \tag{3}$$ but I can not find the procedure to obtain those relations (Actually I found a very strange procedure that I didn't understand at all) and I have been trying but I can't solve this problem. Does anybody could help me please?.

$\endgroup$
0
$\begingroup$

$$pV=RT\left [1+\frac{A_2}{V}+\frac{A_3}{V^2}+...\right ]\\\Rightarrow p=RT\left [\frac 1V+\frac{A_2}{V^2}+\frac{A_3}{V^3}+...\right] \\\Rightarrow p^2 = R^2T^2\left [ \frac {1 }{V^2}+ \frac{2A_2}{V^3}+...\right]$$

Now substitute for $p$ and $p^2$ into the equation

$$pV=RT[1+B_2p+B_3p^2+...]$$

and gather up terms is $\dfrac 1V$ and $\dfrac{1}{V^2}$ and then compare them with the first equation.

$\endgroup$
0
$\begingroup$

Hints:

  1. Define a "density/inverse volume" $$\alpha~:=~\frac{RT}{V}, \tag{A}$$ and rescale the $T$-dependent virial coefficient functions $$ A_k^{\prime}~:=~\frac{A_k}{(RT)^{k-1}} . \tag{B}$$

  2. Then OP's 2 virial expansions (1) & (2) read $$p~\stackrel{(1)}{=}~\alpha[1+A^{\prime}_2\alpha+A^{\prime}_3\alpha^2+\ldots] \tag{1'}$$ and $$\alpha~\stackrel{(2)}{=}~p[1+B_2p+B_3p^2+\ldots]^{-1},\tag{2'}$$ respectively.

  3. Next, in eq. (2') use the formulas for reciprocal power series: $$\alpha~\stackrel{(2')}{=}~p[1+B^{\prime}_2p+B^{\prime}_3p^2+\ldots].\tag{2''}$$

  4. Finally eqs. (1') & (2'') are mutually connected via power series reversion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.