How is dense matter related to temperature?

Jun 1, 2016

The less energy there is (this could be heat energy - linking to temperature), the denser the object will be.

Explanation:

As particles gain energy, they start to move more and faster. This leads to the particles being more spread out and therefore less dense.

Therefore, an object with less energy is likely to be denser than one with more. A molecule of water making up solid, dense ice will have less energy than one making up gaseous, less dense steam.

Think of a room of toddlers. If they have no energy, they will all sit together and sleep in a small area. They are densely packed.
If you give them some ice cream and they get lots of energy then they will move around more and spread out.

I hope this helps; let me know if I can do anything else:)

Jun 6, 2016

In a perfect gas under constant pressure, the density is inverse proportional to the temperature.

Explanation:

This question is complicate because the particles behave differently if they are in a gas, in a solid in a superfluid or in one of the many state of the matter.
I will consider only a gas (actually a perfect gas) because the concept of temperature and density are well defined.

For a perfect gas we have the equation:

$P V = {k}_{b} N T$

where $P$ is the pressure, $V$ is the volume, ${k}_{b}$ is a the constant of Boltzmann, $N$ is the number of particles and $T$ is the temperature.

The density can be defined as the number of particles per unit of volume $\setminus \rho = \frac{N}{V}$. If we consider that the pressure stay constant, then we have

$T \frac{N}{V} = \frac{P}{k} _ b$

$T = \frac{K}{\setminus} \rho$

where $K$ is the constant quantity $K = \frac{P}{k} _ b$.

So the temperature is inverse proportional to the density, it means that when the density increases, the temperature decreases and vice-versa.