Unit problem with calculating the mass concentration of a particle

[guitarstorm]

Homework Statement

We have to calculate the mass concentration for a particle size distribution. I'm just doing the first size range here, since once I get that the rest is just repetition.

FYI, the particles are spherical.

Size range (diameter) = 0.005-0.5 μm
Average particle size (diameter) = 0.25 μm
Number concentration = 1 cm$^{-3}$
Particle density = 1800 kg m$^{-3}$

The answer is supposed to be in units μg m$^{-3}$ μm$^{-1}$

Homework Equations

$m=\rho _{p}\frac{\pi }{6}D_{p}^{3}n_{n}(D_{p})$

The Attempt at a Solution

First thing I did was convert the density from kg m$^{-3}$ to μg m$^{-3}$:

$\frac{1800 \: kg}{m^{3}} * \frac{1\: \mu g}{1*10^{9\: }kg}=\frac{1.8*10^{12}\: \mu g}{m^{3}}$

Then I plugged into the mass concentration formula:

$(1.8*10^{12}\mu g\: m^{-3})(\frac{\pi }{6})(0.25\: \mu m)^{3}(1\: cm^{-3})=1.473*10^{10}\mu g\: m^{-3\: }\mu m^{3}\: cm^{-3}$

So now, since there's no cm in the units we're supposed to have, I figured I'd get rid of that:

$(1.473*10^{10}\: \frac{\mu g\: \mu m^{3}}{m^{3}\: cm^{3}})*(\frac{100\: cm}{1\: m})^{3}=1.473*10^{16}\: \frac{\mu g\: \mu m^{3}}{m^{6}}$

So now we have this mess, and at this point I give up, because idk how the heck to get this in μg m$^{-3}$ μm$^{-1}$... Not to mention the number I'm getting seems way too large...

 

[mark726]

Hey there,

Thank you for sharing your attempt at solving this problem. I can see that you have made some progress, but there are a few errors and misunderstandings that I would like to address.

Firstly, when converting the density from kg m^{-3} to μg m^{-3}, you should be dividing by 10^9, not multiplying. This is because 1 kg is equal to 10^9 μg.

So the correct conversion would be:

\frac{1800 \: kg}{m^{3}} * \frac{1\: \mu g}{10^{9\: }kg}=\frac{1.8\: \mu g}{m^{3}}

Next, when plugging in the values into the mass concentration formula, you have the particle density in the wrong place. The formula should be:

m = \rho _{p}\frac{\pi }{6}D_{p}^{3}n_{n}(D_{p})

So the correct calculation would be:

(1.8\: \mu g\: m^{-3})(\frac{\pi }{6})(0.25\: \mu m)^{3}(1\: cm^{-3})=1.767*10^{10}\mu g\: m^{-3}

Next, when converting to the desired units of μg m^{-3} μm^{-1}, you should be dividing by the particle size (0.25 μm), not multiplying. This is because you want the mass concentration per unit of particle size.

So the final calculation would be:

1.767*10^{10}\: \frac{\mu g}{m^{3}}\: \frac{1}{0.25\: \mu m}=7.068*10^{10}\: \frac{\mu g}{m^{3}\: \mu m}

This is very close to the desired units of μg m^{-3} μm^{-1}. To get the exact units, you can use the fact that 1 μm = 10^{-6} m. So you can rewrite the units as:

7.068*10^{10}\: \frac{\mu g}{m^{3}\: \mu m} = 7.068*10^{4}\: \frac{\mu g}{m^{3}\: m^{-1}} = 7.068*10^{4}\: \frac{\mu g}{m^{