What specifically were you wanting? I think that by independently working through the problems in these videos, you'd come away with a pretty solid foundation on the topics he presents.
Sir, at 3:35 you said that the pressure will not change considerably and thus took it as a constant, why is the same not true for volume also? Looking at the extreme left portions of the graph, the change in pressure would be HUGE for a small change in volume. Thanks for the videos.
That statement was in context to the principles of integration (and had nothing to do with physics). When there is an infinitesimal change in the volume (dV) then the change in pressure over that interval is essentially zero.
When the equation is written as dU = Q + W then W is the work done ON the gas When the equation is written dU = Q - W then W is the work done BY the gas.
i ve got a question. in this isothermic situation, you'd say delU =0 so that Q = W. but, in the previous videos; you'd say Q = n.c.del T; in this situation del T = 0. How does Q not equal zero?
+Metehan Yetiştiren Using the first law of thermodynamics: delta U = Q - W. Therefore if delta U is zero because it is a isothermic process and the temperature is constant, then Q = W.
Hi, thank you very much for your videos. It's very instructif..please I have a question: why did you say that (delta U= n C delta T ) because we know that Q which equal = nc delta T , consequently in this case that T = constant Q which should be equal o zero and delta U= Q+ W so delta U = W and not zero.. please clarify to me thanks
In an isothermic process ( T = constant), Q is not zero. (Q is only zero in an adiabatic process). Since T is constant, delta U MUST be zero. If delta U is zero and delta U = Q - W, then W = Q.
thank's for your answer. Actually I thought that since T= constant =T2=T1 so delta T = T2-T1=0 SO Q=O ISN'T IT? AND I DIDN'T GET HOW DELTA U = n C delta T AND Q THE SAME THING SO HOW DELTA U = Q+W ?THANKS
No, it just means that all the heat added to the gas is used by the gas to do work and none of it is used to change the internal energy of the gas. (That is why T is constant and delta U is zero).
I don't quite understand why we cannot use W=P*delta V, I know you did say P is changing constantly but how do we know it from the graph? or is it depending on the constant temperature?
+Attiyah Lanin You can use it if you can figure out how the pressure changes. (Like with a isothermal process). It is better to use the specific equations for work done by the gas for each type of thermodynamic process.
Yes. To put this in an example with units: 2 bars * 6 liters = 3 bars * 4 liters The bars are a unit of pressure (each is 100 kPa), roughly equal to one atmosphere. Liters would be the unit of volume
I have a question. I am confused with the statement, "work is done by increasing heat. But the temperature does not change." How is it possible that even though heat is added, temperature is constant?
Thank you so much man. I've really been struggling and it just clicked.
Thanks a lot! This explanation made everything fit together in my head!
You are the best Mr!!!! Thanks a lot.
HELLO SIR!! I referred to your website but there was not anything extra/different that what is here
That is correct.
@@MichelvanBiezen Sir do you have any book or handout related to this lessons?
Not yet. We are considering making those in the future.
What specifically were you wanting? I think that by independently working through the problems in these videos, you'd come away with a pretty solid foundation on the topics he presents.
Thank you sir. Very good videos. I definitely like how you show the P V diagram, it really helped me understand the process better.
Sir, at 3:35 you said that the pressure will not change considerably and thus took it as a constant, why is the same not true for volume also? Looking at the extreme left portions of the graph, the change in pressure would be HUGE for a small change in volume. Thanks for the videos.
That statement was in context to the principles of integration (and had nothing to do with physics). When there is an infinitesimal change in the volume (dV) then the change in pressure over that interval is essentially zero.
First of all thanks for the quick reply...... But then shouldn't the work done dW = d(PV) and then by chain rule, dW = VdP + PdV ???? Thanks again.
This is very exciting video I would like to ask do you have videos on Carnot cycle?
Yes we do. (And more to come).
@@MichelvanBiezen please share the link because I'm desperately in need of it right now
Try searching his channel or his playlists.
AWESOME!!! I'M ACING MY EXAM NOW!!!
Did you comment that while taking the exam? 😃
Hi, this video is brilliant. For the change in internal energy, does it make a difference using Q+W? Thanks for this video.
When the equation is written as dU = Q + W then W is the work done ON the gas When the equation is written dU = Q - W then W is the work done BY the gas.
Ok, thank you.
i ve got a question. in this isothermic situation, you'd say delU =0 so that Q = W. but, in the previous videos; you'd say Q = n.c.del T; in this situation del T = 0. How does Q not equal zero?
+Metehan Yetiştiren
Using the first law of thermodynamics: delta U = Q - W. Therefore if delta U is zero because it is a isothermic process and the temperature is constant, then Q = W.
+Metehan Yetiştiren that was in isovolumetric not isothermic
I love it when things cancel out or are zero. ΔU = 0 ftw!
Makes things a lot easier!
do you have two different playlists for thermodynamics: states?
There are summary playlists as well as the individual "chapter" playlists for physics. (they contain the same videos)
okay, thank you. I love your videos.
6:45 thank god
Hi, thank you very much for your videos. It's very instructif..please I have a question: why did you say that (delta U= n C delta T ) because we know that Q which equal = nc delta T , consequently in this case that T = constant Q which should be equal o zero and delta U= Q+ W so delta U = W and not zero.. please clarify to me thanks
In an isothermic process ( T = constant), Q is not zero. (Q is only zero in an adiabatic process). Since T is constant, delta U MUST be zero. If delta U is zero and delta U = Q - W, then W = Q.
thank's for your answer. Actually I thought that since T= constant =T2=T1 so delta T = T2-T1=0 SO Q=O ISN'T IT? AND I DIDN'T GET HOW DELTA U = n C delta T AND Q THE SAME THING SO HOW DELTA U = Q+W ?THANKS
No, it just means that all the heat added to the gas is used by the gas to do work and none of it is used to change the internal energy of the gas. (That is why T is constant and delta U is zero).
I don't quite understand why we cannot use W=P*delta V, I know you did say P is changing constantly but how do we know it from the graph? or is it depending on the constant temperature?
+Attiyah Lanin You can use it if you can figure out how the pressure changes. (Like with a isothermal process). It is better to use the specific equations for work done by the gas for each type of thermodynamic process.
P1V1 = P2V2. So for example, if P1V1 = 2 * 6 = 12 then P2V2 can possibly equal 3 * 4 = 12 right? Just an example.
Yes. To put this in an example with units:
2 bars * 6 liters = 3 bars * 4 liters
The bars are a unit of pressure (each is 100 kPa), roughly equal to one atmosphere.
Liters would be the unit of volume
I think we can take integral in terms of pressure instead of volume. Are there any specific reasons for the integral of volume, not pressure?
You can do either one. Very good observation.
@@MichelvanBiezen Thank you for your answer!
I have a question. I am confused with the statement, "work is done by increasing heat. But the temperature does not change." How is it possible that even though heat is added, temperature is constant?
All of the heat added is coverted to work. (None of it is used to heat up the gas)
@@MichelvanBiezen Thank you for your answer professor!
I suppose you could imagine the volume increasing just enough so that the temperature never rises...