# Thermodynamics - Change in Density due to Change in Height

1. Feb 20, 2010

### Hart

1. The problem statement, all variables and given/known data

A column of water contains fine metal particles of radius 20nm, which are in thermal
equilibrium at 25°C.

The density of the metal is $$2\times10^{4} kg m^{-3}$$.

If there are 1000 particles per unit volume at given height, what will the particle density per unit volume be at a position of 1 mm higher?

2. Relevant equations

Stated within the question / solution attempt.

3. The attempt at a solution

For each particle:

Radius = $$20\times10^{-9}m$$

Hence Volume = $$\frac{4}{3} \pi (20\times10^{-9}m^{3} = 3.351\times^{-23}m^{3}$$

Therefore since density is mass divided by volume, $$\rho = \frac{m}{V}$$:

$$m_{total} = \rho m = (2\times10^{4} kg m^{-3})(3.351\times^{-23}m^{3}) = 6.702\times^{-19}kg$$ which is the total contained mass.

It is given that at any height there are 1000 particles, therefore:

$$m_{particle} = \frac{6.702\times^{-19}kg }{1000} = 6.702\times^{-22} kg$$

I thought this would probably help somehow, not sure how anymore though.

Last edited: Feb 20, 2010
2. Feb 21, 2010

### gabbagabbahey

Ask yourself, 'What's keeping the metal particles from falling to the bottom?'.

3. Feb 21, 2010

### Hart

.. little more help?

4. Feb 22, 2010

### Hart

1. Using known radius of a particle have found the volume of a particle.
2. Using this value of volume, and the known pressure, have found the mass.
3. Using this value of mass, and known values of g and h, have found the energy:

$$E = mgh = (m)(9.81)(1\times10^{-3}) = 6.6\times10^{-21}$$

4. Using Boltzmann constant $$k_{B}$$ and T (converted into Kelvin), can now put all these values into the Stefan-Boltzmann equation:

$$n = n_{0}exp\left(\frac{-E}{k_{B}T}\right) = (1000)(0.202) = 202$$

5. Therefore at position of 1mm higher, the particle density is: 202 particles per unit volume.

.. hopefully correct