# Thread: Determing an Equation Problem

1. ## Determing an Equation Problem

This problem is in relation to multivariable local linearization, but I can't seem to understand what the problem is asking for!

Problem:

The coefficient, β, of thermal expansion of a liquid relates the change in the volume V (in m3) of a fixed quantity of a liquid to an increase in its temperature TC) as shown below.

(a) Let ρ be the density (in kg/ m3) of water as a function of temperature. Write an expression for dρ in terms of ρ and dT. You can use rho for ρ and beta for β.

What I've Tried/Done To Solve It:

I did some research online about the relationship between density and temperature as well as Volume and temperature in terms of thermal expansion. (I'm assuming you shouldn't have to do this research to solve the problem, but I thought it might help me). I learned that Volume and temperature are directly proportional. I also learned that density and temperature are inversely proportional. At least I *think* this is true, and I'm not exactly sure how it would all relate to thermal expansion. Anyway, I noticed that the equation involving volume above showed a directly proportional relationship, using β as a sort of proportionality constant.Based on that, I guessed that the relationship between density and temperature would look similar, but inversely proportional, like this:

d
ρ = /ρ) dT

I would really appreciate any help, even if it's just to push me in the right direction. Explanations of anything you're doing would be wonderful. I'd really like to learn conceptually what these types of problems are asking for. Thank you!

2. "Volume is directly proportional to temperature" for a ideal gas (pV= nRT). That is NOT true for a liquid. Also, for either, it is true that volume will increase with increasing temperature so it should be clear that density will decrease. Your formula, $d\rho= \frac{\beta}{\rho}dT$ has $d\rho$ positive.

You are told that $dV= \beta VdT$. Density is defined by $\rho= \frac{m}{V}= mV^{-1}$ where m is the (constant) mass. Differentiate that to get $d\rho= mV^{-2}dV= -\frac{mdV}{V^2}$ and use that with $dV= \beta VdT$

3. From first DE you can find V(T).
$\rho=m/V$
where m=const

4. Originally Posted by HallsofIvy
"Volume is directly proportional to temperature" for a ideal gas (pV= nRT). That is NOT true for a liquid. Also, for either, it is true that volume will increase with increasing temperature so it should be clear that density will decrease. Your formula, $d\rho= \frac{\beta}{\rho}dT$ has $d\rho$ positive.

You are told that $dV= \beta VdT$. Density is defined by $\rho= \frac{m}{V}= mV^{-1}$ where m is the (constant) mass. Differentiate that to get $d\rho= mV^{-2}dV= -\frac{mdV}{V^2}$ and use that with $dV= \betaV dT$
So based on what you've explained, this is what I did:

Knowing that
$d\rho = \frac{-m}{V^2}$

and

$dV = \beta V dT$

I solved the first equation for dV and set the two equal, like this:

$\frac{-V^2}{m} d\rho = (\beta * V) dT$

Solving for $d\rho$, I got this:

$d\rho = \frac{m\beta*V}{-V^2} dT$

$d\rho = \frac{m\beta}{-V} dT$

So would this then be correct?

$d\rho = -\rho\beta dT$