Thermal Expansion of a Liquid

[EEristavi]

Homework Statement

A liquid has a density $\rho$.
Show that the fractional change in density for a change in temperature $\Delta$T is
$\frac {\Delta \rho} {\rho} = -\beta \Delta T$

Relevant Equations

$\Delta V = \beta V \Delta T$

$\Delta V = \beta V \Delta T$$\rho _2 = \frac m {V + \Delta V} = \frac { \rho V} {V(1+\beta \Delta T)} = \frac \rho {1+\beta \Delta T}$
$\Delta \rho = \rho _2 - \rho = \frac \rho {1+\beta \Delta T} - \rho = \rho (\frac 1 {1+\beta \Delta T} - 1) = \rho \frac {-\beta \Delta T} {1+\beta \Delta T}$

$\frac {\Delta \rho} {\rho} = \frac {-\beta \Delta T} {1+\beta \Delta T}$

I can't figure out were I make the mistake...
It should be:
$\frac {\Delta \rho} {\rho} = -\beta \Delta T$

 

[haruspex]

Homework Statement: A liquid has a density $\rho$.
Show that the fractional change in density for a change in temperature $\Delta$T is
$\frac {\Delta \rho} {\rho} = -\beta \Delta T$
Homework Equations: $\Delta V = \beta V \Delta T$

$\Delta V = \beta V \Delta T$$\rho _2 = \frac m {V + \Delta V} = \frac { \rho V} {V(1+\beta \Delta T)} = \frac \rho {1+\beta \Delta T}$
$\Delta \rho = \rho _2 - \rho = \frac \rho {1+\beta \Delta T} - \rho = \rho (\frac 1 {1+\beta \Delta T} - 1) = \rho \frac {-\beta \Delta T} {1+\beta \Delta T}$

$\frac {\Delta \rho} {\rho} = \frac {-\beta \Delta T} {1+\beta \Delta T}$

I can't figure out were I make the mistake...
It should be:
$\frac {\Delta \rho} {\rho} = -\beta \Delta T$

There is no mistake, you just need to take the first order approximation.
The ΔT in the denominator is insignificant compared with the 1 in the denominator.

 

[EEristavi]

There is no mistake, you just need to take the first order approximation.
The ΔT in the denominator is insignificant compared with the 1 in the denominator.

$\Delta T$ or $\beta?$
Because as I think $\Delta T$ Can me "big"

Also,
Can you give me some direction: where I can look it up - Why I can discard that value in denominator?
I know I will find it eventually, but I would appreciate it if you could help me spare some time

P.S. I kind of tried to expand this as a Tailor (maclaurin) Series, but was't satisfied with the answer..

 

[haruspex]

Because as I think ΔT Can me "big"

It can be, but the expression you are asked to derive for change in density is not valid for large changes in temperature.
E.g. consider a temperature gain so great that the volume doubles. βΔT=1. What does that give for the new density, according to the target formula?

 

[mfig]

I think an easy approach is to simply differentiate the fundamental relationship between volume, mass and density.

$m = \rho V\tag{1} \rightarrow V = m \rho^{-1}$

If we differentiate this for a unit of mass:

$dV = -m\rho^{-2} d \rho \tag{2}$

Now simply plug (1) and (2) into the relation

$d V = \beta V d T \tag{3}$

Your desired equation immediately results.

 

[EEristavi]

What does that give for the new density, according to the target formula?

It would be impossible to use first order approximation - so answer will be "my formula"?

 

[haruspex]

It would be impossible to use first order approximation - so answer will be "my formula"?

The question should have specified that βΔT is small. You are expected to use the first order approximation.

 

[EEristavi]

Thanks all