Thermodynamics - Change in Density due to Change in Height

[Hart]

Homework Statement

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?

Homework Equations

Stated within the question / solution attempt.

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.

 

[gabbagabbahey]

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

 

[Hart]

.. little more help?

 

[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

 

[paranormal]

Firstly, it is important to note that the density of water changes with temperature. At 25°C, the density of water is approximately 997 kg/m^3.

Assuming that the metal particles are evenly distributed throughout the water column, the density of the mixture will be a combination of the density of water and the density of the metal particles. This can be calculated using the following formula:

\rho_{mixture} = \frac{m_{water} + m_{particles}}{V_{water} + V_{particles}}

Where:
m_{water} = mass of water
m_{particles} = mass of metal particles
V_{water} = volume of water
V_{particles} = volume of metal particles

Since the temperature and pressure are constant, the volume of water will not change as we move up the column. However, the volume of the metal particles will change with height due to the change in pressure.

Using the ideal gas law, we can calculate the change in pressure with height:

P_{2} = P_{1}e^{-\frac{Mgh}{RT}}

Where:
P_{2} = pressure at height h
P_{1} = pressure at initial height (25°C)
M = molar mass of water (18.02 g/mol)
g = acceleration due to gravity (9.8 m/s^2)
h = height difference (1 mm = 0.001 m)
R = gas constant (8.314 J/mol K)
T = temperature (25°C = 298 K)

Plugging in these values, we get:

P_{2} = 1\times10^{5} Pa (approximately)

Now, using the ideal gas law again, we can calculate the volume of the metal particles at the new pressure:

V_{2} = \frac{nRT}{P_{2}}

Where:
V_{2} = volume at new pressure
n = number of moles of metal particles
R = gas constant (8.314 J/mol K)
T = temperature (25°C = 298 K)
P_{2} = pressure at new height

Since we know the number of particles per unit volume (1000 particles per unit volume), we can calculate the number of moles of particles present at the new height:

n_{2} = \frac{1000 particles}{V_{2}}

Finally, we can calculate the new volume of the metal particles:

V