# Repost/Merged Threads - Density: pressure and temperature dependency

• Tales Ferraz
In summary: Also, what is your goal in solving for β? Is it just to get an equation that describes how β changes with P, or is there some other reason you want to know β?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'.Also, what is your goal in solving for β? Is it just to get an equation that describes how β changes with P, or is there some other reason you want to know
Tales Ferraz
Homework Statement
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

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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.

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

Tales Ferraz
Tales Ferraz said:
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?

Last edited:
Tales Ferraz
haruspex said:
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!

mitochan said:
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

Last edited:
mitochan said:
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
$$d\rho=(\frac{\partial \rho}{\partial T})_p dT +(\frac{\partial \rho}{\partial p})_T dp$$

mitochan and Tales Ferraz
haruspex said:
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!

Tales Ferraz said:
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'.

Tales Ferraz

## 1. What is the relationship between density and pressure?

The relationship between density and pressure is an inverse one. As pressure increases, the volume of a substance decreases, leading to an increase in its density. This is known as Boyle's Law.

## 2. How does temperature affect the density of a substance?

Temperature and density have a direct relationship. As temperature increases, the particles within a substance have more energy and move faster, causing the substance to expand and become less dense. Conversely, as temperature decreases, the substance contracts and becomes more dense.

## 3. Why does air density decrease with altitude?

Air density decreases with altitude because as altitude increases, there is less air above pushing down on the air below. This decreases the pressure, causing the air molecules to spread out and the air to become less dense.

## 4. How does density impact the behavior of fluids?

Density plays a crucial role in the behavior of fluids. Heavier fluids, or those with a higher density, will sink below lighter fluids. This is why oil, which has a lower density than water, will float on top of water. Additionally, the density of a fluid can affect its ability to flow and its resistance to pressure and temperature changes.

## 5. What is the difference between density and specific gravity?

Density is a measure of how much mass is contained in a given volume of a substance. Specific gravity, on the other hand, is a ratio of the density of a substance to the density of a reference substance, typically water. It is a unitless number that is useful for comparing the densities of different substances.

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