Calculating Molecular Diameter: A Guide

[new90]

Homework Statement

How can I calculate diameter of a molecule

Relevant Equations

none

How can I calculate diameter of a molecule

 

[BvU]

find a relationship between something with a known number of molecules and its volume

 

[new90]

thanks, but the issue is that I don't have the number of molecules because I that's the answer of the problem

 

[hutchphd]

perhaps you should state the entire problem rather than us playing "twenty questions"...?

 

[new90]

ok,sorry...

 

[new90]

About 1860 Maxwell sowed that the viscosity of a gas is given by η=ρνι where ρ iπs density ν mean velocity and ι mean free path.The latter quantity he had earlier shown to be ι = 1/(square root of 2)(π)(Ng)(d2)where d is diameter of a molecule.Loshmidt(1865)used the measured value of η,p(gas) and p(solid) together with Joules calculation of ν to determine Ng the number of molecules per cm3 in a gas at STP(1atm 0°C).He assumed the molecules to be hard spheres tightly ppacked ina solid.Given η = 2*10^-4 cm^-1 s^-1 for air at STP ρ(liquid)≈ 1 g cm^-3 p(gas)≈1*10^-3 g cm-3 ,and ν≈ 500 m/s-1. Calculate Ng(number of molecules)

 

[BvU]

So, what are you missing ?

 

[new90]

the diameter of the molecule

 

[BvU]

He assumed the molecules to be hard spheres tightly ppacked ina solid

How are you going to use this ?

 

[DaveE]

Why do you think they gave you two density values, one for liquid, and one for gas?

 

[new90]

i know because it it tighly packed the density of liquid and solid are very similar

 

[DaveE]

yes, so how can you use that information?

 

[new90]

in the problem has a sentence that said "Loshmidt(1865)used the measured value of η,p(gas) and p(solid) together with Joules calculation of ν to determine Ng the number of molecules per cm3 in a gas at STP(1atm 0°C)".

 

[new90]

hello?

 

[DaveE]

So your problem is to find d, and clearly that has something to do with the densities given, which are for air. Would the answer be different for H₂ instead of air?

Notice that the density given includes mass (g) in the dimensions, but mass doesn't appear anywhere else in the equations. This needs to be resolved somehow with external information.

Also note that there is a typo as presented you (or they) have mixed up the gas density and the liquid density in the problem. Easy to resolve though, use the larger one for liquids, the smaller one for gas.

 

[DaveE]

Here's a related problem that may help:
A 1 cm x 1cm x 1cm cube of Aluminum has a density of 2.70 g/cm3. How many Al atoms are there in the cube? How far apart are they?

 

[new90]

Here's a related problem that may help:
A 1 cm x 1cm x 1cm cube of Aluminum has a density of 2.70 g/cm3. How many Al atoms are there in the cube? How far apart are they?
what is the answer?

 

[new90]

I don't know how to get the Al atoms

 

[DaveE]

OK, a 1cm x 1cm x 1cm cube has a volume of 1 cm³. With a density of 2.70 g/cm³, this means there are 2.70 g of Al in the cube. So, how many Al atoms weight 2.70 g? What information do you need to get to solve this problem? Where/how will you get it?

Then, when you find out how many atoms there are. How far apart are they (center to center, since they will be touching each other)? Imagine carefully placing tennis balls inside a big box to get as many as possible inside.

Finally, for the original problem, you will need to know that air is composed of 80% N₂ and 20% O₂. A N₂ molecule weighs twice as much as a N atom.

 

[new90]

thnaks for the help we solve it

 

[DaveE]

Congratulations, good work!