# Thermal Expansion of a Liquid

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 Helper
Gold Member
2020 Award
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
Homework Helper
Gold Member
2020 Award
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?

Last edited:
• TSny and DEvens
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}##

• 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