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.
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
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?
Hi there! I'd be happy to help you tackle this problem.
To find the mass of 1 L of air, you need to consider the composition of air, which is 80% nitrogen (N2) and 20% oxygen (O2) by volume. However, we need the mass of air, so we need to convert from volume to mass. Since air is a mixture of gases, we can assume ideal gas behavior and use the ideal gas law to convert volume to mass.
First, we can calculate the molar mass of air by taking into account the molar masses of nitrogen and oxygen. Nitrogen (N2) has a molar mass of approximately 28 g/mol, and oxygen (O2) has a molar mass of approximately 32 g/mol. The molar mass of air can be calculated using the weighted average of these values:
Molar mass of air = (0.8 * 28 g/mol) + (0.2 * 32 g/mol) = 28.8 g/mol
Next, we can use the ideal gas law, PV = nRT, to relate the volume of air to its mass. Assuming 1 atm pressure, 25 degrees Celsius (298 K) temperature, and a molar mass of air of 28.8 g/mol:
PV = nRT
(1 atm) * V = (m/M) * (R * T)
V = (m/M) * (R * T)
Here, V represents the volume of air (1 L), m is the mass of air, M is the molar mass of air, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
Now, we can rearrange the equation to solve for m (mass of air):
m = (V * M * P) / (R * T)
m = (1 L * 28.8 g/mol * 1 atm) / (0.0821 L·atm/(mol·K) * 298 K)
m ≈ 1.209 g
So, the mass of 1 L of air is approximately 1.209 g.
To determine the mass of CCl4 in the air, we need to convert the concentration of CCl4 from ppm (parts per million) by mass to grams.
Since 25 ppm by mass means that for every million parts of air, there are 25 parts of CCl4, we can calculate the mass of CCl4 (m_ccl4) in 1 L of air using the following equation:
m_ccl4 = (25 g / 1,000,000 g) * (1.209 g)
m_ccl4 ≈ 0.030 g
Now that we have the mass of CCl4 in 1 L of air, we can calculate the partial pressure (P_ccl4) of CCl4 using Dalton's law of partial pressures. Considering that the total pressure (P_total) is 1 atm:
P_ccl4 = (m_ccl4 / M_ccl4) * P_total
P_ccl4 = (0.030 g / 153.8 g/mol) * 1 atm
P_ccl4 ≈ 0.000195 atm
Therefore, the partial pressure of CCl4 in the lab atmosphere is approximately 0.000195 atm.
I hope this helps clarify the steps to solve the problem. Let me know if you have any further questions!