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# PV-diagrams and expansion work

Why work from expansion is the area under the curve of a PV-diagram. Why heat is not a state function and internal energy is a state function.  Created by Sal Khan.

## Want to join the conversation?

• I'm confused. Around the 2 minute mark you say we're assuming gradual incremental change, because otherwise the value of P for pressure would decrease. But won't it decrease anyway? Even if the change in volume were infinitesmally small, wouldn't that just create an equally small change in pressure? Thus when delta V goes up P always goes down? So it's less P by more dV. If the drop in P and the rise in delta V are proportional, then how can the value of W increase with increasing volume? AARGH! • Your question is 8+ years old, and my answer is most probably not relevant to you, but it may be relevant to future readers.
In the video, Sal didn't actually ignore the ∆P as the volume changed. He just took the ∆P in incremental steps instead of constantly taking its change, and we can clearly see in the PV diagram that Sal didn't ignore the ∆P because if he had done so, our area under the curve would've been a simple quadrangular.
Instead, Sal simply went down in the pressure for every new beginning of a new quasistatic process, kinda like how you go down in stairs, "abruptly", and not in a "smooth" way, like in a slide.
But why did Sal choose to"go down in stairs" instead of a "smooth slide"? Well, recall that Sal proved that W=Px∆V, and he used that fact to show the viewer that Work = area under the curve in a PV diagram.
So, he had to use stairs, because in each individual stair, the pressure is constant, in order to show us, the viewrs, that Work = the area under the curve.
Now, if we use smaller and smaller "stairs" to describe this process, i.e. infinitesimally small quasistatic processes, we practically get a smooth slide, just like how the actual process should be. Sal initially used "bigger" stairs to drive home the point that
Work = Area under the curve in a PV diagram, that's all there's to it.
• Isn't the work being done by whoever took the rock off of the piston? Cuz rocks don't just disappear. • It doesn't make sense to me.
We're taking change in pressure(dP) to be negligible and thus take P in the formula as constant. At the same time, we're taking the (dV) change in volume as significant. When we're taking very small increments, how can we consider only dV and ignore dP? If dP is negligible, so is dV!
Also, when the graph is made, pressure is shown to change! • Robby's is a good answer (and the question is a good one, too). For each bar, we find the area by multiplying the TOTAL internal pressure by the CHANGE in volume, hence P and deltaV. But for each subsequent interval (bar) we do changge the pressure to a new value. We are not pretending that pressure is constant, we allow it to change from step to step, but the quantity we use is still total pressure (P), rather than deltaP.
• After completing a full cycle, a certain amount of work is produced by the system. If the internal energy (U) of the system remains constant from the beginning of the cycle until it reaches the end, where does the produced work (energy) come from? Generating work without altering the energy of the system (U) seems like energy is generated from nothing. • its a fact that no perfect ideal gas exists..a gas may have similar characters..so if there is no ideal gas..what is the use of studying about it? • The ideal gas law turns a very complicated problem of solving a system with trillions of particles into a simple algebraic equation with the penalty of losing a small amount of accuracy. If you want to make the problem much more difficult to gain a small amount of additional accuracy, you are welcome to do so.
• At the very end, Sal said that the change in U for ideal gases was 0. For the graph that had two different paths, both for fowards and backwards states, how does this apply since the amount of work done was different? • I hope I'm answering this question (wat too late)
But an ideal gas is defined as a gas where the molecules is so far apart that they dosn't react to eachother. This leads to that the internal energy for an ideal gas is only depending on Q (not pressure or volume). only heat.
Therefore U is still the same even if we change V or P
(I hope i got it right, maybe someone could fill in)
• Why does Sal draw the PV-diagrams as linear functions (y=ax+b)? The equation behind the PV-diagram is P*V=n*R*T, so shouldn't the diagram by hyperbolic since n*R*T stays constant in these scenarios? • Sal says PV=constant ..... hence P = constant / V ...so shouldn't the graph be like a curve ? • Isn't the inaccuracy in finding the work in each measured area coming from the fact that you are only measured the area of the rectangle instead of including the triangle above it as well? Assuming that the pressure changes form a straight line, in order to get an accurate value, couldn't you get the area of the rectangle, then use an equation to find the area of a right triangle to find the area above, then add both to find the correct area?
(1 vote) • I suggest that you watch the calculus videos, it will help to make sense of finding the areas under curves. You are right, when dV is not infinitely small, then it is not exactly a perfect rectangle. But using calculus, we can make it so that we are adding up all the rectangles where dV is so infinitely small that the little triangle at the top doesn't exist. I'm sure Sal can explain this a lot better than me, in his Calculus videos :)
• Also, i still dont get the idea why W=(deltaP)(deltaV) shouldnt be true.
and why W=P(deltaV) is true.
many say deltaP is negligible, but thats the same case for volume isnt it?
since its a quasi static process, volume and pressure both increase in infinitesimal increaments.
hence, if we account for deltaV why dont we account for the corresponding deltaP

thankyou 