(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]2\times10^{4} kg m^{-3}[/tex].

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 = [tex]20\times10^{-9}m[/tex]

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

Therefore since density is mass divided by volume, [tex]\rho = \frac{m}{V}[/tex]:

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

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

[tex]m_{particle} = \frac{6.702\times^{-19}kg }{1000} = 6.702\times^{-22} kg[/tex]

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

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# Homework Help: Thermodynamics - Change in Density due to Change in Height

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