How Do You Derive β for the Given Partial Differential Equation?

[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 don't know the latex format for pd

 

[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.

 

[j-lee00]

Cheers

 

[coomast]

You're welcome.