Hey Fiona, you're intuition is correct, the real gas pressure will be less than the ideal. That is why I have P = Pi - Pa where P is the real pressure, Pi is the ideal pressure, and Pa is the pressure contribution of the attractive forces. Essentially what we do is solve this equation for "Pi" and then substitute that into the ideal gas expression, so it becomes Pi = P + Pa. Now we have an equation that deals with the real pressure of the system. Meanwhile the term that is being subtracted from the Volume, is correcting the real volume directly. We are subtracting the excluded volume term from the real volume. This is a little difficult to explain just using text but hopefully that clarifies it a bit.
Hello, not sure what I am doing wrong, but I cannot seem to get the answer for B) part ii. Is there any canceling out of values happening that I may be missing. I keep getting approximately the same answer as I did for the Ideal gas answer for A) part ii
Hi! Greetings from Argentina. I'm a new fan of you and your channel! I've tried so hard to find a demonstration of where does the pressure correction come from but, I couldn't. I would appreciate if you can help me to understand whether that was just empirical or not. Maybe an article or sth like that. Thanks a lot!
wow this is just saving me. Thank you so much Professor! I have never thought that chemistry can be so easy to understand
If the pressure is decreased in real gases, why is it (p + correction) rather than (p - correction) as is the case with volume?
Hey Fiona, you're intuition is correct, the real gas pressure will be less than the ideal. That is why I have P = Pi - Pa where P is the real pressure, Pi is the ideal pressure, and Pa is the pressure contribution of the attractive forces. Essentially what we do is solve this equation for "Pi" and then substitute that into the ideal gas expression, so it becomes Pi = P + Pa. Now we have an equation that deals with the real pressure of the system.
Meanwhile the term that is being subtracted from the Volume, is correcting the real volume directly. We are subtracting the excluded volume term from the real volume. This is a little difficult to explain just using text but hopefully that clarifies it a bit.
Thank you, that was very useful
Hello, not sure what I am doing wrong, but I cannot seem to get the answer for B) part ii. Is there any canceling out of values happening that I may be missing. I keep getting approximately the same answer as I did for the Ideal gas answer for A) part ii
thank you for what you do
Please in the second question where we have 1000k, what does the difference in our calculated pressure tells us?
The unit of a seems to be wrong, it should be dm6 atm mol-2.
perfect bro
Hi! Greetings from Argentina. I'm a new fan of you and your channel! I've tried so hard to find a demonstration of where does the pressure correction come from but, I couldn't. I would appreciate if you can help me to understand whether that was just empirical or not. Maybe an article or sth like that. Thanks a lot!
Why are your units for VDW coefficients so strange?
For a I find L^2 bar mol^-2. You had... dm^6 atm mol^-1. dm3 is 1L , so why L^2 and why mol^ -1?