Visit ilectureonline.com for more math and science lectures! In this video I will explain the molar heat capacity of a gas.
Пікірлер: 73
@winniejeng74024 жыл бұрын
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.
@historyisthebest58314 жыл бұрын
I think that he has been teaching for at least 7 - 8 years since it's 2020 now.
@terrytan43307 жыл бұрын
Your teaching is outstanding sir. Thank You
@1stcomment7004 ай бұрын
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. KZfaq University truly is the best.
@MichelvanBiezen4 ай бұрын
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.
@1stcomment7004 ай бұрын
@@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.
@MichelvanBiezen4 ай бұрын
Thank You!
@slurp11136 жыл бұрын
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!
@farazshams39649 жыл бұрын
You are going to get me through engineering school, I am watching these videos to get a jump start come fall! Thank you
@ronthedon87304 жыл бұрын
He didn't say "Welcome to Ilecture online!" :O
@cafikes2511 жыл бұрын
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!
@colorx60302 жыл бұрын
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.
@MichelvanBiezen2 жыл бұрын
Glad it helped!
@hanaanbread4 жыл бұрын
Now I can actually understand the formula rather than mindless memorization, so THANK YOU! :D
@kalinzlatilov20507 жыл бұрын
Great explanation. 1 recommendation only: Put all this into practice with real examples. Cheers.
@ledung71605 жыл бұрын
This is true gem
@md.mominulislam50688 жыл бұрын
sir, i had many problems about the concept of PV diagram. But you solved my problems. Thanks a lot.
@nambirajannambi56275 жыл бұрын
Thank you
@jbergamp2 ай бұрын
Thanks u
@MichelvanBiezen2 ай бұрын
You are welcome.
@vogotikrishna82703 жыл бұрын
From INDIA🇮🇳🇮🇳🇮🇳
@adetayoeyelade79247 жыл бұрын
extremely great teaching!
@jamessontimola11348 жыл бұрын
Thank you mr. Thermo you are awesome sir
@vusimusawenkosimahlangu54263 жыл бұрын
Thank you so much.
@tiwarykhan74376 жыл бұрын
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
@musicislife6658 жыл бұрын
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?
@MichelvanBiezen8 жыл бұрын
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.
@musicislife6658 жыл бұрын
Thank you and I wanted to say that I really appreciate your lessons.
@stephanosphilippou13007 жыл бұрын
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
@JanHalda8 жыл бұрын
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?
@MichelvanBiezen8 жыл бұрын
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
@JanHalda8 жыл бұрын
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.
@MichelvanBiezen8 жыл бұрын
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
@rubixcubesolve7 жыл бұрын
But for case 3, Isn't Q=0 when temperature dosen't change?
@greetingsfromromania48857 жыл бұрын
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.
@user-xm4yd5jv2x3 ай бұрын
I am hearing you saying we learn the equation here , how can i find those videos
@MichelvanBiezen3 ай бұрын
Which equation are you referring to?
@shaherbano13347 жыл бұрын
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?
@MichelvanBiezen7 жыл бұрын
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.
@shaherbano13347 жыл бұрын
Got it Sir. Thanks
@warmscar2368 жыл бұрын
Could you tell me,What different between Molar Heat Capacity and Specific Heat Capacity?
@MichelvanBiezen8 жыл бұрын
+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.
@warmscar2368 жыл бұрын
Michel van Biezen Thank you.
@krishnaawatar25856 жыл бұрын
very nice sir
@sidhusandevamanoharan95367 жыл бұрын
@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
@MichelvanBiezen7 жыл бұрын
They are just constants. Like A + B = C. No errors here.
@sidhusandevamanoharan95367 жыл бұрын
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?"
@MichelvanBiezen7 жыл бұрын
If you watch the other videos on this topic on this channel it will become clear.
@agentbrickschannel45234 жыл бұрын
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...
@MichelvanBiezen4 жыл бұрын
That is correct, especially when delta V = 0. Then the work done = 0.
@agentbrickschannel45234 жыл бұрын
@@MichelvanBiezen but that ecuation is used when preassure is constant, and there it's not
@MichelvanBiezen4 жыл бұрын
It doesn't matter if the change in volume = 0 No work can be done if there is no change in volume.
@agentbrickschannel45234 жыл бұрын
@@MichelvanBiezen ooooh greaat! Thank you so much!!!! Keep it up! Great videos!
@sigxtizero24414 жыл бұрын
Hi my teacher tells me that Cv for polyatomic gas is 6/2R is that correct? And what does polyatomic gas mean?
@MichelvanBiezen4 жыл бұрын
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)
@sigxtizero24414 жыл бұрын
@@MichelvanBiezen Thank you
@somanguuu9 жыл бұрын
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
@MichelvanBiezen9 жыл бұрын
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
@somanguuu9 жыл бұрын
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...
@MichelvanBiezen9 жыл бұрын
Melody Tak Cv is the specific heat of a gas at constant volume
@RiyaTomar-jd3mw4 жыл бұрын
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 ?
@MichelvanBiezen4 жыл бұрын
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.
@RiyaTomar-jd3mw4 жыл бұрын
@@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
@abdulrahmanmohamed82985 жыл бұрын
which value is the R value? is it 0.0821 or 8.314?
@MichelvanBiezen5 жыл бұрын
Both. It depends on which units you use: R = 8.314 J/mol K (standard units) or R = 0.0821 liter atm /mol K
@nemanjazivaljevic16569 жыл бұрын
There is work by constant pressure and by constant volume!?But it seems that you only use work by constant pressure,why?Thanks!
@MichelvanBiezen9 жыл бұрын
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
@nemanjazivaljevic16569 жыл бұрын
Michel van Biezen When yo have change in pressure you have work W=Vdp but in this example you write it as zero.Why?
@nemanjazivaljevic16569 жыл бұрын
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?
@satyamkumar5674 жыл бұрын
U r god
@nijansikrakar79899 жыл бұрын
what if p and v both changes plz help sir michel
@MichelvanBiezen9 жыл бұрын
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.