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right this is the question. at a temperature of 300K and 1atm the density of air is 1.29kgm^-3 and the co-eff of viscosity is 1.75x10^-5Pas. Air has a molecular weight of 29 and the average speed is given by the square root of 8kt/piem.
what is their mean free path.
so i have two equations main equations to use
1/(4sqrt(2)pi x n x d^2 and the other is kt/sqrt(2) x pi x d^2 x pressure
so i use the equation that pv=nRT to find the volume (is that correct or could i use the density = mass/volume) when i have found the volume i then equate it to 4/3pi r^3 and re arrange for r and then multiply it by 2 to get the diameter. I then use the second equation but it doesn't come out correct (i have a list of possible answers and i just can't seem to get them though). Do you convert the weight into kg by dividing through by 1000 or can you just leave it as 29 atomic masses. (which would be 29 multiplied by 1.66x10^-27)?
any suggestions on where iv gone wrong would be greatly appreciated.
thanks
adam
what is their mean free path.
Homework Equations
so i have two equations main equations to use
1/(4sqrt(2)pi x n x d^2 and the other is kt/sqrt(2) x pi x d^2 x pressure
The Attempt at a Solution
so i use the equation that pv=nRT to find the volume (is that correct or could i use the density = mass/volume) when i have found the volume i then equate it to 4/3pi r^3 and re arrange for r and then multiply it by 2 to get the diameter. I then use the second equation but it doesn't come out correct (i have a list of possible answers and i just can't seem to get them though). Do you convert the weight into kg by dividing through by 1000 or can you just leave it as 29 atomic masses. (which would be 29 multiplied by 1.66x10^-27)?
any suggestions on where iv gone wrong would be greatly appreciated.
thanks
adam