For a case of gas to liquid, does the slope of the curve approximates to 0 because gas particles become liquid and don't increase the pressure? and is the line left to the coexistence curve imaginary? because there's no gas anymore.
1) Yes, as the molar volume decreases (and the density increases) in the flat area the pressure stays constant as any increase in density results in more particles becoming liquid, freeing up more space for the remaining gas molecules as the liquid phase clumps together tightly 2) The line to the left of the coexistence curve is real, but it is the phase where all of the molecules are now liquid. At this point, very small increases in density result in dramatic increases in pressure, as liquids are highly incompressible relative to gases. If you increase the pressure on a liquid you can decrease its volume a tiny bit, but not nearly as much as you could for a gas. This is the basis of hydraulic pumps, as enormous amounts of work can be done by applying pressure to water, as rather than compress the water will change shape in the pipe, performing some useful work along the way.
Hi! Could you clarify something for me? Won't the first and second derivatives of pressure with respect to volume both equal zero at multiple temperatures? But there is only one temperature where the critical point exists. So those derivatives alone don't define the critical point?
Hi Jami. One thing we know about the pressure vs. molar volume graph is that it must be monotonically decreasing. This means that at no point can we increase the pressure while increasing the molar volume. Though there are plenty of locations where increasing the molar volume keeps a constant pressure, due to a change of some percentage of the contents from the liquid phase to the gas phase. At low temperatures, this happens over a wide molar volume change, at slightly higher temperatures it is smaller, and at high temperatures above some threshold it doesn't happen at all. So temperatures above the critical temperature Tc, the first derivative of P vs. V is always negative. Below Tc it discontinuously jumps to zero, stays there a while, and then discontinuously jumps again to negative. But at the critical point, and only at the critical point, the first derivative just barely goes up to zero for an instant, then becomes negative again. So in one sense you could say it's the point where the first derivative continuously goes to zero. But if we then look at the ideal gas equal and other corrections to it like the van der Waals equation of state, we notice that they form a Taylor series for P in terms of powers of 1/V (ideal gas is 1st order, vdw is 3rd order). Ideal gas can't model this, because if the derivative is ever non-zero negative, it's always non-zero negative. Vdw however, is a cubic equation and can model this. We can set the 1st derivative equal to zero at a given point. But this isn't sufficient, since the derivative of a 3rd order function is a 2nd order function, which can have multiple points where the 1st derivative is zero, unless we apply an additional constraint. If we apply the constraint that the 1st derivative must equal zero somewhere, *and* that the 1st derivative must never be positive, we get a fully constrained (i.e. solvable) equation, which tells us what given the critical properties of a gas (Tc, Pc, Vc) our choice of parameters a and b must be in order to satisfy those constraints. That's not quite a full answer, but that's a slightly deeper examination of the relevant factors up to the state of my knowledge.
Hey, Thank you for the videos. I'm doing Thermodynamics and Kinetics this semester and they've been wonderful. At your subcritical isotherm (T2), you said "the pressure will start going up with volume" I think you probably meant pressure becomes constant or something. You want to clarify this?
Really good video! Thank you.. One question: Can the difference in the critical point be explained by the differences of the van der walls parameters (a) and (b)?
Not sure exactly what you mean, Bangali. If you're referring to the analogous video that has a "1" in the lower left corner of the thumbnail, the difference is that is the old version of this video, and this is the new one. I've completed updated almost every video on this channel, and the new versions have a "2" in the lower left corner of the thumbnail. The old ones also have {old version} in the thumbnail and/or title.
this 7 minute video taught me more than my professor could in 2 days.
you're an amazing teacher!
Happy to help, Soumil. Keep up the good work.
Thank you for this video! Perfect!
For a case of gas to liquid, does the slope of the curve approximates to 0 because gas particles become liquid and don't increase the pressure?
and is the line left to the coexistence curve imaginary? because there's no gas anymore.
1) Yes, as the molar volume decreases (and the density increases) in the flat area the pressure stays constant as any increase in density results in more particles becoming liquid, freeing up more space for the remaining gas molecules as the liquid phase clumps together tightly
2) The line to the left of the coexistence curve is real, but it is the phase where all of the molecules are now liquid. At this point, very small increases in density result in dramatic increases in pressure, as liquids are highly incompressible relative to gases. If you increase the pressure on a liquid you can decrease its volume a tiny bit, but not nearly as much as you could for a gas. This is the basis of hydraulic pumps, as enormous amounts of work can be done by applying pressure to water, as rather than compress the water will change shape in the pipe, performing some useful work along the way.
@@TMPChem I wanted to understand the liquid pressure part. Thanks
Hi! Could you clarify something for me? Won't the first and second derivatives of pressure with respect to volume both equal zero at multiple temperatures? But there is only one temperature where the critical point exists. So those derivatives alone don't define the critical point?
Hi Jami. One thing we know about the pressure vs. molar volume graph is that it must be monotonically decreasing. This means that at no point can we increase the pressure while increasing the molar volume. Though there are plenty of locations where increasing the molar volume keeps a constant pressure, due to a change of some percentage of the contents from the liquid phase to the gas phase. At low temperatures, this happens over a wide molar volume change, at slightly higher temperatures it is smaller, and at high temperatures above some threshold it doesn't happen at all. So temperatures above the critical temperature Tc, the first derivative of P vs. V is always negative. Below Tc it discontinuously jumps to zero, stays there a while, and then discontinuously jumps again to negative. But at the critical point, and only at the critical point, the first derivative just barely goes up to zero for an instant, then becomes negative again. So in one sense you could say it's the point where the first derivative continuously goes to zero. But if we then look at the ideal gas equal and other corrections to it like the van der Waals equation of state, we notice that they form a Taylor series for P in terms of powers of 1/V (ideal gas is 1st order, vdw is 3rd order). Ideal gas can't model this, because if the derivative is ever non-zero negative, it's always non-zero negative. Vdw however, is a cubic equation and can model this. We can set the 1st derivative equal to zero at a given point. But this isn't sufficient, since the derivative of a 3rd order function is a 2nd order function, which can have multiple points where the 1st derivative is zero, unless we apply an additional constraint. If we apply the constraint that the 1st derivative must equal zero somewhere, *and* that the 1st derivative must never be positive, we get a fully constrained (i.e. solvable) equation, which tells us what given the critical properties of a gas (Tc, Pc, Vc) our choice of parameters a and b must be in order to satisfy those constraints. That's not quite a full answer, but that's a slightly deeper examination of the relevant factors up to the state of my knowledge.
How did you jump to the values of the three critical quantities?
Hey, Thank you for the videos.
I'm doing Thermodynamics and Kinetics this semester and they've been wonderful.
At your subcritical isotherm (T2), you said "the pressure will start going up with volume" I think you probably meant pressure becomes constant or something.
You want to clarify this?
The pressure will increase dramatically as it becomes an incompressuble fluid
Really good video! Thank you.. One question: Can the difference in the critical point be explained by the differences of the van der walls parameters (a) and (b)?
Your question should be the other way around.
Thank you,but can't find the difference between video 1 & 2.
Not sure exactly what you mean, Bangali. If you're referring to the analogous video that has a "1" in the lower left corner of the thumbnail, the difference is that is the old version of this video, and this is the new one. I've completed updated almost every video on this channel, and the new versions have a "2" in the lower left corner of the thumbnail. The old ones also have {old version} in the thumbnail and/or title.
Oh sorry, I didn't get it then.
I thought the videos will show different method of derivation.
Both have been very helpful, thank you so much
Thanks for watching