Holy moly you answered questions in the comment section from 3 years ago, 4 months ago, 2 years ago, and 2 months ago, and this video is uploaded in 2013? You're the real stuff.
What a masterpiece. I graduated engineering and took thermodynamics 4 years ago but I feel like I didn't learn anything and barely scraped by the course. But this series covers every single detail without overloading you with TOO many random details and I feel like I'm actually understanding the fundamentals of thermodynamics now. TH-cam University truly is the best.
First of all congratulations for graduating with your engineering degree, that is quite an accomplishement. Glad you found our videos to rekindle you interest in the subject.
@@MichelvanBiezen Thank you for your great work. I am using this channel to study for the FE exam. I just finished this entire 1st Law of Thermodynamics playlist. Your style of teaching made everything seem so much simpler than I remember in school. Donations coming your way.
These videos are amazing. I thought I was going to fail my physics final tomorrow since my engineering lectures are hard to understand and the teacher is not very good, but coming across your channel saved me. Everything is very nicely paced and well explained. Thank you so much!
I'm confused at 4:55 where you take Q = nCvdeltaT and apply it to the State 1 --> State 3 equations. Doesn't Q = nCvdeltaT only apply at constant volume, whereas in the case of State 1 to State 3 the volume is changing and the pressure is constant? How can you use Cv and Cp in the same equation? Thanks!
Oh wow, that is much clearer. Thanks a lot! Our source material derives the equations using the letter d instead of the triangle symbol so perhaps that's why it gets confusing to me. Changing the letter d (derivative) to Delta symbol makes all the difference.
I don't understand why in 2:31 you use the W=P. Delta V to describe the work done by the transformation from 1 to 2, if the preassure in that change is not constant...
@5:00 a Cv and Cp are being added. How is it possible to add those. One dictates the change in internal energy with constant volume and ther dictates with constant pressure
Michel van Biezen Hello sir . First of all thank you for the reply. Now to the question. we use the constant Cv when the system has a constant volume throughout process and we use the constant Cp when the system has a constant pressure throughout the process . I dont understand how in a isobaric process both these constants are used in the first law equation. My question is "shouldn't there be another constant instead of Cv in isobaric process to dictate the amount of internal energy?"
That is only for Q (the heat added to or removed from the gas). The equation for delta U is always = n Cv delta T. The internal energy of a gas CANNOT depend on how it got there.
I firstly want to say that your videos are deeply appreciated.If you could further explain this question that would be great Why is it that ΔU is always using Cv as it's constant even in the examples of a changing Volume ? I understand the solution under this assumption i just don't get the intuition behind it Thanks in advance
May I kindly ask one thing I don't understand about your explanation. In the second case where gass changes from STATE 1 to STATE 3 you use nCp∆T=Q. You took the ∆U=Q from the first case (STATE 1 to STATE 2), where Q=∆U, because ∆V=0→W=0. But you cannot use that in the same way with isobaric change, where W≠0, because there is a change in volume (∆V≠0). If you do the same derivation of Q in the second case, you don't get Q=nCp∆T. Or what am I missing?
In the first case we go from state 1 to state 2. (constant volume). Therefore W = 0 and Q = nCvdeltaT = delta U (I didn't use delta U = nCpdeltaT because that is never true) When the pressure is constant Q = nVpdelta T and delta U = nCvdeltaT
Yes, I agree, and understand everything about first case, but in the second one you used Q = nCp∆T. All I don't understand is why you used this. I mean I don't understand where Q = nCp∆T is taken from. And from the way you derivated the Q from an isobaric gass change it looks like you used it same way in the isochoric one, only with Q changed form nCv∆T to nCp∆T.
Nope, my friend, when you add heat into an isothermic process all the heat is used not to increse the internal energy, but it's converted into work. So all the heat will be used to expand the gas, and not to increase temperature, so yes, delta T = 0, but that doesn't mean the energy added is 0, there is heat added, but it's "used for another purpose", if i may say that.
Sir, can you please explain me that why diatomic molecules dont exhibit vibrational motion? why only translational & rotational when they are given energy via heat?
Diatomic molecules can vibrate, but due to the quantum mechanic restrictions of the molecule, it will require a large jump in energy absorption to start it vibrating at the first vibrational quantum state, thus under normal circumstances it does not add to the heat energy of the molecule.
+Warm Scar Specific heat capacity is used in terms of mass as applied to solids and liquids. The molar heat capacity is used for gases and it does depend on the type of molecules in the gas.
Polyatomic means there are 3 or more atoms in the molecule. The actual value vary quite a bit for polyatomic molecules, but the typical value used in most text books will be 7/2 R (again there is no exact value and if your teacher uses 6/2 R that is OK)
Melody Tak SPECIFIC HEAT is the amount of heat needed to heat a specific quantity of gas (1 mole) by 1 degree C or K HEAT CAPACITY is the amount of heat needed to heat the total amount of gas in consideration by 1 degree C or K Heat capacity is often defined as the ratio of the heat added divided by the increase in temperature and will be different for different quantities of gas
Michel van Biezen I saw somewhere, you said you use Physics for Scientists and Engineers.. in ch 21 on the title and subtitles, they definite Cv as specific heat.. So the book is wrong...
Nemanja, I am not sure what you are asking, but this may clear it up: Change in internal energy delta U = n*Cv*delta T (always) Heat added to the gas (Q) depends on the process; Q = n * Cv * delta T if the volume remains constant Q = n * Cp * delta T if the pressure remains constant
Nijan, Cv and Cp are constants, so they don't change. The change in internal energy of a gas doesn't depend on how the gas changes. It only depends on the change in temperature. The work done by a gas can be found by integrating PdV. You must find P as a function of V and then integrate over the change in volume which is the area under the curve.
sir,the relation between Cp and Cv is- Cp - Cv = R for any gas . now there's a question . its given that Cp - Cv =a for H2 and Cp - Cv = b for N2. relation between a and b is asked. Shouldn't it be R only???? cuz this is the wrong answer and correct answer is a = 14b. HOW??? Why the molar mass is taken into account ?
Cv and Cp are defined as being calculated per mol and therefore the molar mass is not relevant. So if you read the question carefully, there may be something there to indicate you need to take the molar mass into account. But it is either poor practice or wrong by the author of the problem.
@@MichelvanBiezen sir , this is the complete question that came in JEE 2017. Another same type of question that came in JEE 2007 says - If Cp and Cv are specific heats of N2 per unit mass then relation between Cp and Cv is - again , the answer should be Cp-Cv=R as given in one of the four options but the answer is Cp-Cv=R/28 .......................... SIR PLEASE HELP
Holy moly you answered questions in the comment section from 3 years ago, 4 months ago, 2 years ago, and 2 months ago, and this video is uploaded in 2013? You're the real stuff.
I think that he has been teaching for at least 7 - 8 years since it's 2020 now.
Don't you mean, "holy mole-y"? 😎
What a masterpiece. I graduated engineering and took thermodynamics 4 years ago but I feel like I didn't learn anything and barely scraped by the course. But this series covers every single detail without overloading you with TOO many random details and I feel like I'm actually understanding the fundamentals of thermodynamics now. TH-cam University truly is the best.
First of all congratulations for graduating with your engineering degree, that is quite an accomplishement. Glad you found our videos to rekindle you interest in the subject.
@@MichelvanBiezen Thank you for your great work. I am using this channel to study for the FE exam. I just finished this entire 1st Law of Thermodynamics playlist. Your style of teaching made everything seem so much simpler than I remember in school. Donations coming your way.
Thank You!
@@1stcomment700 Out of curiosity, how'd the FE exam go? Do you currently work in engineering?
Your teaching is outstanding sir. Thank You
These videos are amazing. I thought I was going to fail my physics final tomorrow since my engineering lectures are hard to understand and the teacher is not very good, but coming across your channel saved me. Everything is very nicely paced and well explained. Thank you so much!
That's awesome! How'd the final go?
I'm confused at 4:55 where you take Q = nCvdeltaT and apply it to the State 1 --> State 3 equations. Doesn't Q = nCvdeltaT only apply at constant volume, whereas in the case of State 1 to State 3 the volume is changing and the pressure is constant? How can you use Cv and Cp in the same equation?
Thanks!
You are going to get me through engineering school, I am watching these videos to get a jump start come fall!
Thank you
Nine years later, did engineering school go how you hoped it would?
Oh wow, that is much clearer. Thanks a lot! Our source material derives the equations using the letter d instead of the triangle symbol so perhaps that's why it gets confusing to me. Changing the letter d (derivative) to Delta symbol makes all the difference.
Glad it helped!
I don't understand why in 2:31 you use the W=P. Delta V to describe the work done by the transformation from 1 to 2, if the preassure in that change is not constant...
That is correct, especially when delta V = 0. Then the work done = 0.
@@MichelvanBiezen but that ecuation is used when preassure is constant, and there it's not
It doesn't matter if the change in volume = 0 No work can be done if there is no change in volume.
@@MichelvanBiezen ooooh greaat! Thank you so much!!!! Keep it up! Great videos!
From INDIA🇮🇳🇮🇳🇮🇳
How are things going in India these days?
Great explanation. 1 recommendation only: Put all this into practice with real examples. Cheers.
@5:00 a Cv and Cp are being added. How is it possible to add those. One dictates the change in internal energy with constant volume and ther dictates with constant pressure
They are just constants. Like A + B = C. No errors here.
Michel van Biezen Hello sir . First of all thank you for the reply. Now to the question.
we use the constant Cv when the system has a constant volume throughout process and we use the constant Cp when the system has a constant pressure throughout the process . I dont understand how in a isobaric process both these constants are used in the first law equation. My question is "shouldn't there be another constant instead of Cv in isobaric process to dictate the amount of internal energy?"
If you watch the other videos on this topic on this channel it will become clear.
He didn't say "Welcome to Ilecture online!" :O
Binge-watching these videos, that phrase had become engrained in my brain!
Now I can actually understand the formula rather than mindless memorization, so THANK YOU! :D
But the volume is not constant when you go from 1 till 3 so why do you use deltaU = n cv deltaT? cv is only when the volume is constant or not?
That is only for Q (the heat added to or removed from the gas). The equation for delta U is always = n Cv delta T. The internal energy of a gas CANNOT depend on how it got there.
Thank you and I wanted to say that I really appreciate your lessons.
I firstly want to say that your videos are deeply appreciated.If you could further explain this question that would be great
Why is it that ΔU is always using Cv as it's constant even in the examples of a changing Volume ? I understand the solution under this assumption i just don't get the intuition behind it
Thanks in advance
May I kindly ask one thing I don't understand about your explanation.
In the second case where gass changes from STATE 1 to STATE 3 you use nCp∆T=Q. You took the ∆U=Q from the first case (STATE 1 to STATE 2), where Q=∆U, because ∆V=0→W=0. But you cannot use that in the same way with isobaric change, where W≠0, because there is a change in volume (∆V≠0). If you do the same derivation of Q in the second case, you don't get Q=nCp∆T. Or what am I missing?
In the first case we go from state 1 to state 2. (constant volume). Therefore W = 0 and Q = nCvdeltaT = delta U
(I didn't use delta U = nCpdeltaT because that is never true) When the pressure is constant Q = nVpdelta T and delta U = nCvdeltaT
Yes, I agree, and understand everything about first case, but in the second one you used Q = nCp∆T. All I don't understand is why you used this. I mean I don't understand where Q = nCp∆T is taken from. And from the way you derivated the Q from an isobaric gass change it looks like you used it same way in the isochoric one, only with Q changed form nCv∆T to nCp∆T.
Let's review Q for the 4 processes:1) Isovolumetric: (V= constant) Q = nCvdeltaT2) Isobaric: (P = constant) Q = nCpdeltaT3) Isothermic (T = constant) Q = W (delta U = 0)4) Adiabatic Q = 0and Cp = Cv + R
But for case 3, Isn't Q=0 when temperature dosen't change?
Nope, my friend, when you add heat into an isothermic process all the heat is used not to increse the internal energy, but it's converted into work. So all the heat will be used to expand the gas, and not to increase temperature, so yes, delta T = 0, but that doesn't mean the energy added is 0, there is heat added, but it's "used for another purpose", if i may say that.
sir, i had many problems about the concept of PV diagram. But you solved my problems. Thanks a lot.
Sir, can you please explain me that why diatomic molecules dont exhibit vibrational motion? why only translational & rotational when they are given energy via heat?
Diatomic molecules can vibrate, but due to the quantum mechanic restrictions of the molecule, it will require a large jump in energy absorption to start it vibrating at the first vibrational quantum state, thus under normal circumstances it does not add to the heat energy of the molecule.
Got it Sir. Thanks
Oh man, these lectures are hot 🔥🔥🔥
This is true gem
We use constant k for molecules and R for moles ...can you plzz explain me why you use contant R for molecules instead of K
Could you tell me,What different between Molar Heat Capacity and Specific Heat Capacity?
+Warm Scar
Specific heat capacity is used in terms of mass as applied to solids and liquids. The molar heat capacity is used for gases and it does depend on the type of molecules in the gas.
Michel van Biezen Thank you.
extremely great teaching!
I am hearing you saying we learn the equation here , how can i find those videos
Which equation are you referring to?
Hi my teacher tells me that Cv for polyatomic gas is 6/2R is that correct? And what does polyatomic gas mean?
Polyatomic means there are 3 or more atoms in the molecule. The actual value vary quite a bit for polyatomic molecules, but the typical value used in most text books will be 7/2 R (again there is no exact value and if your teacher uses 6/2 R that is OK)
@@MichelvanBiezen Thank you
Thank you mr. Thermo you are awesome sir
Thank you
Thank you so much.
Thanks u
You are welcome.
Is it molar HEAT CAPACITY of a gas or molar SPECIFIC HEAT of a gas? Because Physics by Serway says that it's the molar specific heat of gas
Melody Tak
SPECIFIC HEAT is the amount of heat needed to heat a specific quantity of gas (1 mole) by 1 degree C or K
HEAT CAPACITY is the amount of heat needed to heat the total amount of gas in consideration by 1 degree C or K
Heat capacity is often defined as the ratio of the heat added divided by the increase in temperature and will be different for different quantities of gas
Michel van Biezen I saw somewhere, you said you use Physics for Scientists and Engineers.. in ch 21 on the title and subtitles, they definite Cv as specific heat.. So the book is wrong...
Melody Tak
Cv is the specific heat of a gas at constant volume
which value is the R value? is it 0.0821 or 8.314?
Both. It depends on which units you use: R = 8.314 J/mol K (standard units) or R = 0.0821 liter atm /mol K
very nice sir
There is work by constant pressure and by constant volume!?But it seems that you only use work by constant pressure,why?Thanks!
Nemanja,
I am not sure what you are asking, but this may clear it up:
Change in internal energy delta U = n*Cv*delta T (always)
Heat added to the gas (Q) depends on the process;
Q = n * Cv * delta T if the volume remains constant
Q = n * Cp * delta T if the pressure remains constant
Michel van Biezen When yo have change in pressure you have work W=Vdp but in this example you write it as zero.Why?
Nemanja Zivaljevic Its called technical work and it was used to find enthalpy but it doesen't do any actual work so it is maybe not needed?
what if p and v both changes plz help sir michel
Nijan,
Cv and Cp are constants, so they don't change.
The change in internal energy of a gas doesn't depend on how the gas changes. It only depends on the change in temperature.
The work done by a gas can be found by integrating PdV. You must find P as a function of V and then integrate over the change in volume which is the area under the curve.
sir,the relation between Cp and Cv is-
Cp - Cv = R for any gas . now there's a question . its given that Cp - Cv =a for H2 and Cp - Cv = b for N2. relation between a and b is asked. Shouldn't it be R only???? cuz this is the wrong answer and correct answer is a = 14b. HOW??? Why the molar mass is taken into account ?
Cv and Cp are defined as being calculated per mol and therefore the molar mass is not relevant. So if you read the question carefully, there may be something there to indicate you need to take the molar mass into account. But it is either poor practice or wrong by the author of the problem.
@@MichelvanBiezen sir , this is the complete question that came in JEE 2017. Another same type of question that came in JEE 2007 says - If Cp and Cv are specific heats of N2 per unit mass then relation between Cp and Cv is -
again , the answer should be Cp-Cv=R as given in one of the four options but the answer is Cp-Cv=R/28 .......................... SIR PLEASE HELP
U r god