Finding the pressure required when given number of molecules.

[miniradman]

Homework Statement

To what pressure must a piece of equipment be evacuated in order that there be only 10^8 molecules per ml at 25 degrees celcius

Homework Equations

Ideal gas equation = PV = nRT
Charles's Law = (V2/T2 = V1/T1)

Avagadro's number (6.02 * 10^23)

I know that at STP one mole of gas is 22.4L

The Attempt at a Solution

I don't know where to begin but what I'm trying to do is find how many moles are in 10^8 molecules. Then use the Ideal Gas Equation to figure out the volume at STP and then using Charles's Law, figure out the pressure at 25 degrees or 298K.

 

[hyddro]

ok first you convert the number of molecules to moles using avogadros number.
if you know that 1 mole = 6.02 * 10^23 molecules
you need to find 'x' moles for 10^8 molecules
Once you find the number of moles (which should be a tiny number) convert mL to L (have volume)
You have Temperature (convert to Kelvin)
With this you can find Pressure using the ideal gas equation assuming that R (the constant) = 0.082 L atm K−1 mol−1
Good luck

 

[miniradman]

when I divide 10^8 by 6.02*10^23 I get and outrageously large number, and when I go vice versa, I still get a massive number.

 

[hyddro]

I get a extremely small number when i divide 10^8 by 6.02 *10^23 ... are you doing this correctly? check your steps ..

 

[miniradman]

oh wait, I forgot to put brackets in my calculator :P

but these are my steps. ( I did this at STP)

PV = nRT

101300V = (1.66*10^-16)(8.314)(273)
101300V =

101300V/101300 = 3.767*10-13/101300

v =

P₁/V₁ = P₂/V₂

101300/3.179*10^-18 = P₂/1*10^-3 (BECAUSE ITS 1ML)

the this leads to 2.73*10^16 which is nowhere near the right answer.

the real answer is 4.1*10^-7pa, but I don't know how to get it

 

[hyddro]

you don't have to do it at STP because you are given a specific temperature, you only use STP when you are told to do so... in this case...
T= 25 celsius (convert to Kelvin)
and your pressure is what they are asking you... so P is your unknown...
your volume would be 0.001 L (1ml) ; read carefully the problem 'what pressure [...] in order that there be only 10^8 molecules per ml ..that means 10^8 molecules (or 1.66*10^-16 moles) in only 1 ml (which equals 0.001 L) .. so having all these (plus constant R = 0.082 L atm K−1 mol−1 ) you can solve for P using ideal gas equation...

post your results good luck

 

[miniradman]

Hmmm, I'm getting close.

P(0.001) = (1.66*10^-16) (8.314) (298)
P = 4.11*10^-10

This is close but the answer is ^-7. Am I missing something here?

 

[hyddro]

You got it! :D your answer is in kPa( kilo pascal) 1 kPa = 10^3 Pa, so if you multiply your answer by 10^3 you get what you are supposed to... this comes from the fact that you chose R to be 8.314 and its units are L kPa K−1 mol−1 (note the kPa) ... good job! good luck

 

[miniradman]

but... 8.314 is Pa (according to my textbook)

 

[hyddro]

wait what? can you post the units for constant?

 

[miniradman]

oh wait... the base unit for volume was m^3 so that means that volume isn't 0.001 its 0.000001 which leads to the right answer...

I can't thank you enough, you're awesome dude, live long and prosper \m/

 

[hyddro]

Dude you are welcome! Good luck mate ...\m/