I think that he has been teaching for at least 7 - 8 years since it's 2020 now.

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!

Glad it helped!

You are welcome.

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

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.

Which equation are you referring to?

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

+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.

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.

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!

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

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

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

Both. It depends on which units you use: R = 8.314 J/mol K (standard units) or R = 0.0821 liter atm /mol K

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?

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.