For the average atomic mass of air, use 70 percent N2, and 30 percentO2.
.7*28*+.3*32= it is about 29amu. So you can calculate the mass of one liter, given pressure and temperature.
Now, if you add 25ppm of carbon tet, you have to convert that to amu.
25*molmassCCl4=....amu per million
So divide the avg amu for air by a million, and multipy by the mass in amu for the carbon tet.
Partial pressure= massCCl4/molmass * R*T/V
check my thinking. The actual calcs will require you to use accurate amu's for N2, O2, C, and Cl
Heyy,
Okay so I have a question that I'm trying to figure out how to do and I'm really lost, I was hoping somebody might be able to give me a hand. Thanks in advance.
The question in general is, I know that the vapour pressure of CCl4 is 98mmHg at 25 degrees C. This also means that the equilibrium vapour above liquid CCl4 has the same pressure. We want to know the partial pressure of CCl4 in a lab atmosphere when the CCl4 concentration is 25 ppm by mass. Air is 80% N2 and 20% O2, with the total pressure being 1 atm.
We were given a hint to calculate the mass of say 1 L of air first without CCl4 and then find the mass of CCl4 in it and then calculate the pressure but I still don't get it. I don't know how we would find the mass of 1 L of air...would it just be the mass of O2 and N2 added together? And then if so, when you are saying the CCl4 is 25 ppm by mass...how would you take that away from the mass? Change the mass to ppms as well? Once I get those then I know that you just times the amount of CCl4 that you have by the total pressure to get the partial pressure, but I just can't get that far.
2 answers
I;m going to run through this and see if I can get it and understand it, the only thing I have a question about is why are we changing the percents to 70 and 30?