Repost/Merged Threads - Density: pressure and temperature dependency

[Tales Ferraz]

Homework Statement

Can someone please help me go from equation 2.2 and 2.3 to 2.4a (see image)?

Relevant Equations

α = − (1/ρ)(∂ρ/∂T) (2.2)
β = (1/ρ)(∂ρ/∂P) (2.3)
ρ = ρr.e^β(P−Pr)−α(T −Tr) (2.4a)

See image

 

[Tales Ferraz]

How Beta and Alpha show together in the exponencial? I can solve for only one coefficient (alpha or beta), but not for both. What steps am I missing? Thanks in advance.

 

[mitochan]

Hi. Assuming $\alpha$ and $\beta$ are constant,
$$d\rho=(\frac{\partial \rho}{\partial T})_p dp +(\frac{\partial \rho}{\partial p})_T dp=-\rho \alpha dT + \rho\beta dp$$
$$\frac{d\rho}{\rho}=- \alpha dT + \beta dp$$
$$ln\ \rho - ln\ \rho_r = - \alpha (T-T_r) + \beta( p-p_r)$$

 

[haruspex]

Homework Statement: Can someone please help me go from equation 2.2 and 2.3 to 2.4a (see image)? I mean, how Beta and Alpha show together in the exponencial? I can solve for only one coefficient (alpha or beta), but not for both. What steps am I missing? Thanks in advance.
Homework Equations: α = − (1/ρ)(∂ρ/∂T) (2.2)
β = (1/ρ)(∂ρ/∂P) (2.3)
ρ = ρr.e^β(P−Pr)−α(T −Tr) (2.4a)

See image

Each partial differential equation shows how ρ changes as one of P and T varies but the other stays constant.
You can treat one as an ODE and solve it in isolation, e.g. $\rho=Ae^{-\alpha T}$, but still bearing in mind that this is with the other constant. So A here is a function of the other independent variable, β. Correction: I mean P.
Likewise, $\rho=Be^{-\beta P}$, where B is a function of T.
Can you work it out from there?

 

[Tales Ferraz]

Each partial differential equation shows how ρ changes as one of P and T varies but the other stays constant.
You can treat one as an ODE and solve it in isolation, e.g. $\rho=Ae^{-\alpha T}$, but still bearing in mind that this is with the other constant. So A here is a function of the other independent variable, β.
Likewise, $\rho=Be^{-\beta P}$, where B is a function of T.
Can you work it out from there?

Thank you! I see now the steps I need to make!

 

[Tales Ferraz]

Hi. Assuming $\alpha$ and $\beta$ are constant,
$$d\rho=(\frac{\partial \rho}{\partial T})_p dp +(\frac{\partial \rho}{\partial p})_T dp=-\rho \alpha dT + \rho\beta dp$$
$$\frac{d\rho}{\rho}=- \alpha dT + \beta dp$$
$$ln\ \rho - ln\ \rho_r = - \alpha (T-T_r) + \beta( p-p_r)$$

Thank you so much! But can you show one step before your first step? Like, how did you assumed that dρ=(∂ρ/∂T)pdp+(∂ρ/∂p)Tdp

 

[haruspex]

Hi. Assuming $\alpha$ and $\beta$ are constant,
$$d\rho=(\frac{\partial \rho}{\partial T})_p dp +(\frac{\partial \rho}{\partial p})_T dp$$

That should be
$$d\rho=(\frac{\partial \rho}{\partial T})_p dT +(\frac{\partial \rho}{\partial p})_T dp$$

 

[Tales Ferraz]

Each partial differential equation shows how ρ changes as one of P and T varies but the other stays constant.
You can treat one as an ODE and solve it in isolation, e.g. $\rho=Ae^{-\alpha T}$, but still bearing in mind that this is with the other constant. So A here is a function of the other independent variable, β.
Likewise, $\rho=Be^{-\beta P}$, where B is a function of T.
Can you work it out from there?

Hi again, actually, I'm still stuck. Can you show more steps after the ρ=Ae^−αT, please? I mean, what would you do next. Thank you, in advance!

 

[haruspex]

Hi again, actually, I'm still stuck. Can you show more steps after the ρ=Ae^−αT, please? I mean, what would you do next. Thank you, in advance!

First, see my correction to post #4.
One way would be to differentiate $\rho=A(P)e^{-\alpha T}$ partially wrt P and substitute for $\frac{\partial \rho}{\partial P}$ from your original equations.
Try to get an equation involving only β, A and A'.