[TA1LGUNN3R]

I need to derive α_P and κ_T for a gas obeying Z(ρ,T)=1+B₂ρ+B₃ρ.

κ_T I can find, but I'm having problems with the calculus of α_P.  I know:


[tex]\frac {P}{\rho RT}=1+B_2 \rho+B_3 \rho^2[/tex]

I then isolate for T and take [itex]\frac {\partial T}{\partial \rho}[/itex] and get:

[tex](\frac {\partial T}{\partial \rho})_P [\frac {P}{R}*\frac {1}{\rho+B_2 \rho^2+B_3 \rho^3}]=\frac {-(1+B_2 \rho+B_3 \rho^2)}{(\rho+B_2 \rho^2+B_3 \rho^3)^2}[/tex]

Now, to solve for α_T, I would flip the equation for dρ/dT, and multiply by (-1/ρ), since α_T=[itex]\frac {-1}{\rho}*(\frac {\partial \rho}{\partial T})_P[/itex]

However, I think that looks kinda ugly.  Is this wrong? 

[mjc123]

I think the numerator on the RHS should be -(1+2B₂ρ+3B₃ρ²).
Would it simplify things to equate the denominator to (P/RT)²?

[TA1LGUNN3R]

Yes, in my haste I forgot to type in the coefficients to u'.  I also left out the P/R factor.

Okay, so fixing those, I should get:

[tex](\frac {\partial T}{\partial \rho})_P = \frac {-(T^2 R(1+2B_3 \rho+3B_3 \rho^2)}{P}[/tex]

which implies

[tex]\alpha_P=\frac {P \rho}{T^2 R(1+2B_2 \rho+3B_3 \rho^2)}[/tex]

Doing a unit check, I come up with units of mol²/L² per K.  I guess this sounds right? 

[mjc123]

No, the factor ρ should be on the bottom, then the units are K^-1.

[TA1LGUNN3R]

aargh I don't know why I multiplied by ρ instead of 1/ρ.  Dumb mistake.