It's not even that bad. As long as you set up your limits carefully and do some nice subtraction, you only need to know the integrals of 1/x and (1/x)^gamma. I should do an example sometime.
For an ANSWER that explains it alll a little bit better/slower watch my reaction 2 video's before this one. I 'm not in such a hurry as this Doc always is. He's got a lot to do in his life :-) :-)
Hey, I just recently commented on your other adiabatic process video. And Im still having a bit of trouble understanding adiabats so just to attempt to clarify for myself: In an adiabatic compression... Q=0, therefore deltU = -W, therefore deltU = -PdeltV, therefore Any work done ON the gas (I.e. compression & W 0, deltU must increase meaning temp must increase. And because of ideal gas law, if temp increases, pressure increases. Is this correct? If it is, just a follow up question: how would I use the ideal gas law to justify my last conclusion above? Would I say that because the temp increases and the volume decreases, that causes the pressure to have to increase more than in an isothermal case in which the temp does not increase? So then for an adiabat, the pressure change essentially has to compensate for the decrease in volume AND increase in temp for while isothermal case, the pressure change only has to compensate for volume change. Therefore the pressure for adiabat will be steeper than isotherm, as the video shows?
YES, YES, YES. All of this leads to a more extreme slope for adiabats. You are a great student! The great trouble with IGL for adiabats is that the thing that's held constant (Q) doesn't appear in the lousy equation at all. I think that's why adiabats are the trickiest process for most folks. However, they're also REALLY common in nature (and engines), so we must understand them.
Doc Schuster Awesome! Hopefully I get something on my physics final on Tuesday that has to deal with this concept. Love your videos though! Concise, yet insightful! And I also like how you write out your procedures as opposed to using a computer or powerpoint slides
If you have not learned this trick for watching Physics youtube yet, it is super helpful. Go up to Settings. Go to Playback Speed. Choose a speed that works for you. 0.5 is my go to for all CrashCourse Physics lessons because they are so quick. But really, how fast someone goes is up to you! It is one of the beauties of RUclips.
Good question. Wik: degrees of freedom gas. I don't have a video on this yet, but will surely soon.
Sitting in the library and I just had the biggest Eureka moment thanks to you - Finally understand why temperature can change whilst Q=0
Thanks!!!
Someone elect this man for Congress!
Would that fancy calculus mentioned at 2:52 be double integrals?
It's not even that bad. As long as you set up your limits carefully and do some nice subtraction, you only need to know the integrals of 1/x and (1/x)^gamma. I should do an example sometime.
For an ANSWER that explains it alll a little bit better/slower watch my reaction 2 video's before this one. I 'm not in such a hurry as this Doc always is. He's got a lot to do in his life :-) :-)
n is the number of moles, N is the number of particles.
Aneroid barometer doings how many joules of work
the slope of polytropic process will be more or less steeper than adiabatic process??
@4:15, why 5/2 and 3/2 ._.
Hey, I just recently commented on your other adiabatic process video. And Im still having a bit of trouble understanding adiabats so just to attempt to clarify for myself:
In an adiabatic compression...
Q=0, therefore
deltU = -W, therefore
deltU = -PdeltV, therefore
Any work done ON the gas (I.e. compression & W 0,
deltU must increase meaning temp must increase.
And because of ideal gas law, if temp increases, pressure increases.
Is this correct?
If it is, just a follow up question: how would I use the ideal gas law to justify my last conclusion above?
Would I say that because the temp increases and the volume decreases, that causes the pressure to have to increase more than in an isothermal case in which the temp does not increase?
So then for an adiabat, the pressure change essentially has to compensate for the decrease in volume AND increase in temp for while isothermal case, the pressure change only has to compensate for volume change. Therefore the pressure for adiabat will be steeper than isotherm, as the video shows?
YES, YES, YES. All of this leads to a more extreme slope for adiabats. You are a great student! The great trouble with IGL for adiabats is that the thing that's held constant (Q) doesn't appear in the lousy equation at all. I think that's why adiabats are the trickiest process for most folks. However, they're also REALLY common in nature (and engines), so we must understand them.
Doc Schuster Awesome! Hopefully I get something on my physics final on Tuesday that has to deal with this concept. Love your videos though! Concise, yet insightful! And I also like how you write out your procedures as opposed to using a computer or powerpoint slides
Thank you! I agree - sometimes old technology is more flexible than the systems that replace it.
How does the area under pv curve becomes work done?
Force per distance squared times the change in distance cubed gives you force times distance which is the definition of work
when do we use Delta U= Q-W and Delta U=Q+W?
Depends only on how you define work. Is work positive when the gas does work? This is a simple statement of conservation of energy.
NPH!! at *1.5
I CARE THAT YOU CARE
Why are you so hurry.....
If you have not learned this trick for watching Physics youtube yet, it is super helpful. Go up to Settings. Go to Playback Speed. Choose a speed that works for you. 0.5 is my go to for all CrashCourse Physics lessons because they are so quick. But really, how fast someone goes is up to you! It is one of the beauties of RUclips.
@@rebeccakreidler1893 Maybe he/she means that the video was too short and didn't cover the details. idk