A Partial differetial question

1. Feb 1, 2008

j-lee00

PV = RT(1+B(T)/V)

$$\beta$$ = (1/V)*($$\frac{dV}{dT}$$) at constant P

show $$\beta$$ =$$\frac{1}{T}$$*$$\frac{V + B + T\frac{dB}{dT}}{V + 2B}$$

I got to

$$\beta$$ =$$\frac{PV}{VRT+PRTB}$$*($$\frac{R}{P}$$+$$\frac{d}{dT}$$$$\frac{RTB}{V}$$)

I need help with $$\frac{d}{dT}$$$$\frac{RTB}{V}$$)

I dont know the latex format for pd

Last edited: Feb 1, 2008
2. Feb 3, 2008

coomast

Assuming this is the equation of state (click on any formula in a post to see the latex coding):

$$pV=RT\left(1+\frac{B(T)}{V} \right)$$

The way to proceed is to calculate the derivative

$$\frac{\partial V}{\partial T}$$

By taking the derivative of the equation of state. This gives you:

$$p\frac{\partial V}{\partial T}= R\left(1+\frac{B(T)}{V}\right)+ RT\left(\frac{1}{V}\frac{\partial B(T)}{\partial T}- \frac{B(T)}{V^2}\frac{\partial V}{\partial T}\right)$$

Or:

$$\left[ p+\frac{RTB(T)}{V^2}\right] \frac{\partial V}{\partial T}= R+ \frac{RB(T)}{V}+ \frac{RT}{V} \frac{\partial B(T)}{\partial T}$$

Now the question was to calculate:

$$\beta=\frac{1}{V}\frac{\partial V}{\partial T}$$

Thus you have (after rewriting):

$$\beta=\frac{RT\left(B+V+T\frac{\displaystyle \partial B(T)}{\displaystyle \partial T}\right)}{TV \left(pV+\frac{\displaystyle RTB(T)}{\displaystyle V}\right)}$$

Using now the given equation:

$$\frac{pV}{RT}=1+\frac{B(T)}{V}$$

and substitute in the denominator, you obtain the result asked for.

3. Feb 3, 2008

Cheers

4. Feb 3, 2008

coomast

You're welcome.