Your videos are excellent! I'm really surprised that after 7 years they have so few views, but I suppose not that many people are studying thermodynamics. Thank you so much for doing them! Cheers!
Excellent video! I am no longer a kid (actually, there is a long long long time since i was a kid) and I am fascinated with the ability of this young old man to teach. Congratulations.
Thank you for your comment. We appreciate the positive feedback. I do indeed love teaching and I am glad that through the hard work of my wife (she does all the behind the scene work) I am able to reach out to students and intrested people like yourself around the world. 🙂
Regarding the equation PdV=-nCvdT, how come this assumes that P is kept constant? From my understanding, it should also be affected by a change in temperature in the context given? Is it because we can approximate P to be constant because, at the infinitesimal scale, it would be closer to constant? Or is it due to the fact that P just doesn't change, which is possible if P1V1/T1 = P2V2/T2 and only temperature and pressure change?
@@MichelvanBiezen I understand your point, but there is such a thing as adiabatic irreversible process where there is a net increase in entropy. For instance, a free expansion. A free expansion is adiabatic in that there is no exchange of heat, but it is not isentropic as there is a huge increase in entropy. In a free expansion, the volume would increase, but the temperature does not change, as no work is done.
@@foxhound1008 You are correct. This is a special case of the adiabatic process that is also an isentropic process. Adiabatic processes in general are not necessarily isentropic, since the sudden expansion of a gas in to a vacuum, or a throttling valve process, or an inefficient turbine or compressor, are all examples of processes that are also adiabatic (but not isentropic). If it is adiabatic and reversible, or adiabatic and quasistatic, then it is also isentropic. Isentropic is a special case of adiabatic in general.
@@carultch two other examples of adiabatic, but not isentropic, would be boundary layer flow and flow thru a shockwave. Both are adiabatic, but there is a loss of total pressure, which is a manifestation entropy increase.
Michel a ideal gas (Argon) flows through a turbine and does shaft work (reversible process), why is the change in enthalpy negative? That is H (in) - H (out) < 0....this is a chem E thermo based question. thanks!
Samuel, Under constant pressure delta H = delta U + pV which means that the change in enthalpy equals the change in internal energy plus the work done on the gas. If the gas does work, pV will be negative (in this definition) and thus delta H will be negative (exothermic) The term pV is positive or negative depending HOW it is defined It can be defined as work done BY the gas or work done ON the gas. You have to find out how it id defined in the problem or the book that contains the problem
Yes sir, it is not cool you are stilll using white boards to teach at this stage. You have stuffs but what you write is not clear at all. When i see your videos i just slide pass bevause i know i wont see anything... Make your video animated or always zooom the camera well
Your videos are excellent! I'm really surprised that after 7 years they have so few views, but I suppose not that many people are studying thermodynamics. Thank you so much for doing them! Cheers!
Thanks for watching!
Yeah me too, and I hope more people can find them over the coming years!
Excellent video! I am no longer a kid (actually, there is a long long long time since i was a kid) and I am fascinated with the ability of this young old man to teach. Congratulations.
Thank you for your comment. We appreciate the positive feedback. I do indeed love teaching and I am glad that through the hard work of my wife (she does all the behind the scene work) I am able to reach out to students and intrested people like yourself around the world. 🙂
Sir, you are better than my teacher by miles.
Thank you so much. I am a student of Physics and this really helps a lot!
Glad it was helpful!
How'd the rest of your studies go, or are you still in school?
this helped me so much thanks, perfect within 5mins
This is so clear and easy to follow, thank you!
Thank you sir for teaching so well!
Regarding the equation PdV=-nCvdT, how come this assumes that P is kept constant? From my understanding, it should also be affected by a change in temperature in the context given? Is it because we can approximate P to be constant because, at the infinitesimal scale, it would be closer to constant? Or is it due to the fact that P just doesn't change, which is possible if P1V1/T1 = P2V2/T2 and only temperature and pressure change?
Since this video references an adiabatic process (where P, T, nor V is constant), P is not constant in the PdV = nCvdT equation.
very clear!
Thank you. Glad you think so!
Another good explanation video. Wouldn’t this be for an adiabatic ISENTROPIC process?
An adiabatic process is by definition an isentropic process since there is no exchange of heat.
@@MichelvanBiezen I understand your point, but there is such a thing as adiabatic irreversible process where there is a net increase in entropy. For instance, a free expansion. A free expansion is adiabatic in that there is no exchange of heat, but it is not isentropic as there is a huge increase in entropy. In a free expansion, the volume would increase, but the temperature does not change, as no work is done.
@@foxhound1008 You are correct. This is a special case of the adiabatic process that is also an isentropic process.
Adiabatic processes in general are not necessarily isentropic, since the sudden expansion of a gas in to a vacuum, or a throttling valve process, or an inefficient turbine or compressor, are all examples of processes that are also adiabatic (but not isentropic). If it is adiabatic and reversible, or adiabatic and quasistatic, then it is also isentropic. Isentropic is a special case of adiabatic in general.
@@carultch two other examples of adiabatic, but not isentropic, would be boundary layer flow and flow thru a shockwave. Both are adiabatic, but there is a loss of total pressure, which is a manifestation entropy increase.
Michel a ideal gas (Argon) flows through a turbine and does shaft work (reversible process), why is the change in enthalpy negative? That is H (in) - H (out) < 0....this is a chem E thermo based question. thanks!
Samuel,
Under constant pressure delta H = delta U + pV
which means that the change in enthalpy equals the change in internal energy plus the work done on the gas.
If the gas does work, pV will be negative (in this definition) and thus delta H will be negative (exothermic)
The term pV is positive or negative depending HOW it is defined
It can be defined as work done BY the gas or work done ON the gas.
You have to find out how it id defined in the problem or the book that contains the problem
Gamma? More like "Great!" 👍
Thank you.
Yes sir, it is not cool you are stilll using white boards to teach at this stage. You have stuffs but what you write is not clear at all. When i see your videos i just slide pass bevause i know i wont see anything... Make your video animated or always zooom the camera well
Ok, thanks
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