I just wanted to understand what "adiabatic" means. Watched few other lecturers on YT, couldn't understand anybody. Finally I came here and in the first sentence van Biezen explained it "... when a system does the work but it can't receive energy from outside, it must borrow from it's own internal energy. Reduction of internal energy causes cooling." So simple and elegant one understands whole thermodynamics at once. Thank you!
Thank you so much sir! I fall in love with Physics because of you!!! You are such a great teacher who makes perplexing formulas so easy to understand!!!!!!
Did this many years ago. All professors or lecturers present Thermodynamics differently. They are not trained to be teachers. It's all about you making sure you understand but ask the right questions. Loved the subject.
I just logged in to leave a word of my appreciation. You are a very talented teacher. This channel should get definitely more views. Thank you! You are helping so much!
Thank you so much! This video explained a problem I’ve spent all day on. My teacher is horrible, and doesn’t explain anything, and somehow expected us to derive this without any explanation. Thank you so much. Never stop making videos.
I like to write the "gamma ratios" as γ1, γ2 and γ3, where 1, 2 and 3 are subscripts that represent "monatomic", "diatomic" and "triatomic". I really think that this is the best way to separate them.
Our prof.did not teach us gamma 1.4 value and how it is derived. Thank for detailed explanation. I wish all the best for you. Thank you is small word for this knowledge sharing.
Once agian thank you so much for this lessons dear qualified teacher you honor the profesion of physics lessons ! -Daniel Katana, i can't truly tell you how much i appriciate this art within itself as this year i have chosen Physics as my optional exam, the dear academic professor i appreciate you through philosophy and through the art of anthropology . Respect! You helped me and every student through this pandemic.
Thank you sir for this. I struggled with adiabatic processes bc I don’t know why the formula to calc for T and V is T1V1^c=T2V2^c, where c is Cp/Cv - 1. But now that you showed us how and why the formula got there, it made sense now and it will surely be remembered b/c you taught us how to get there. Thanks a lot sir! This saved so much time to read and understand the derivation in books
Essentially , the process goes so quicly that no heat (or very little heat) will be exchanged with the environment. (like steam escaping from a hole will expand so quickly that it will cool down a lot within a few feet from the hole.
when you write PdV = n Cv dT, the pressure on the left handside is external pressure and let us assume it is constant. The gas does work against external pressure, let us denote it as Pext. But PV=nRT is valid for "internal state" of gas and therefore the P there is internal pressure. If the gas is doing work, P is certainly not equal to Pext (it tries to get there and does work during this process). I doubt if PV=nRT can be used to reduce the entire relation to 2 variables. If this was a reversible process, PV=nRT is perfectly valid. For constant external pressure, I would get dV/dT = n Cv / Pext. Thank you for your time.
I came here lookin for the gas constant because the formula in internet isnt showing its value thanks for the lecture sir! Its very helpful for us reviewee whos not major in mechanical engr
In an adiabatic process, Q = 0, which means that there is no heat exchange with the outside environment. Therefore - W = Change in internal energy ( = dU) And dU is always equal to n Cv dT
This was a great explanation easy to follow along with!! Really appreciate you taking the time out to explain this difficult to understand concept. However, I am extremely new to the physics concept and was hoping you would be able to clarify. I have a question about the 7th going onto the 8th step. Algebraeicaly speaking, shouldn't n from nRT/V(dV)=-nCvdT instead of cancelling out shouldn't that then equal -2n or because it is constant you are able to remove the two n (weight of moles) from the equation?
Mr. Michel, I have found another relationship if I evaluate the limits of the intergral and I assumed the value of the constant to be equal to zero, I obtain: ln [(Tf/To)(Vf/Vo)^(¥ - 1)] = 0 Where ¥ is gamma. Is it correct Mr. Michel?
Prof, if I need to find the constant velocity with which a piston (having mass) moves when it is in an insulated container with a resistor with current flowing through it to generate heat, can I equate the kinetic energy of the piston with the heat generated given that the energy utilised is 60%? Edit: the gas is to be assumed ideal and no volume or pressure terms are a part of the solution which renders the gas laws useless
I know I'm two years late and you have probably already solved this but I'm going to leave this here for anyone who reads this and is wondering. You can't use that approach because since equating kinetic energy to the applied heat would mean that all the work done is being used to accelerate the piston in absence of other forces, but the piston is rising against gravity. Since the piston is rising at a constant velocity the kinetic energy is a constant, but work (heat) is being applied constantly, so this approach would constitute a violation of the conservation of energy. We have to focus on potential energy: the piston is rising with a constant velocity, so the sum of all applied forces is zero. This means that the applied force is equal to the opposite of the weight of the piston, so -mg (we can convert that to simply mg if we assume g as positive). now we equate the supplied heat to the work being done: 0,6IEdt=mgds. ds is the space differential, I current intensity, E charge separation, dt time differential, so Idt is dQ, which is charge differential, and EdQ is the electrical work being done. now the problem tells us that 60% of the electric current is being used to actually lift the weight, so we multiply the electrical work expression by 0,6 and we're set. now we divide everything by mgdt and we obtain ds/dt=0,6IE/mg. ds/dt is the speed (also velocity in this case) so there's your solution.
Wonderful lecture set! Just wondering why in the adiabatic curve, q=mCv delta V only is used, while both temp and pressure change? Not an isovolumetrico or isobaric path at all???? Moves along P and V changing curve. Bij voorbat, bedankt.
Thank you for your help, but it would be nice to put the front Pic that we see before pressing the vedio the as the real quality of vedio pics, so that when we watch we don't feel that this wasn't what we expected!
What is the relationship between temperature, volume and pressure in an adiabatic process? Is it an inverse relationship? I have a statement from a lecture stating as volume decreases, temp and pressure increase. So is it the opposite if volume increases? Thank you for any help on this.
No, you cannot find the integral of zero (nothing). To get a better understanding of what an integral is (or what integration is) and what a derivative is you may want to watch the following playlists: CALCULUS 2 CH 0 WHAT IS INTEGRATION? and CALCULUS 1 CH 2 WHAT IS A DERIVATIVE? at this location: ru-vid.complaylists?view=50&sort=dd&shelf_id=11
Hi , professor. would u like to answer me a question: When you were deriving the formula, you used (C_p = C_v + R), and isn't this formula applicable only under constant pressure conditions? Why could it be used in the current adiabatic process?
im still confused how integration works in physics. For example in the previous video about isothermal you did indefinite integration of dW to W without constant C, but why in this video the indefinite integral of dV/V resulted in lnV with constant C ? I know that indefinite integration always uses C but why not in every case ? Thx
Every indefinite integral has as part of its solution a constant. Depending on the physical situation that constant may be zero or not relevant to the situation.
Sir. when the system is well insulated(adiabatic) i.e Q = 0, does this also mean change in temperature is equal to zero? since Heat depends on T, (Q= mCdT)
No, only during the isothermic process does the temperature remain constant. In an adiabatic process all three, V, T, and P change. When a gas is compressed during an adiabatic process, the work done on the gas causes the temperature to go up (since no heat can escape Q = 0)
If I am understanding it right, change in temperature is used for changing in volume and pressure of internal system but not exchanging heat with surrounding area. Is that would be right statement?
+Sky Lar That is not a silly question at all. It is easier to understand that equation when you compare it to the equation used for a solid:U = m c delta T where m = mass and c = the specific heat.Cv is determined based on the degrees of freedom a gas molecule has and it depends on the shape of the molecule .
+Anes Merazi For deltaU It will always be Cv as the process that's occurring does not matter at deltaU is only dependant of deltaT. Whereas when calculating Q it does matter when you use Cp or Cv. I hope this helps. Someone asked this on the Constant Pressure video (example 1). He explains is there in the comment section.
There's a mistake in that that corrects itself with another mistake, deltaU=CvdT not deltaU=nCvdT, this corrects itself though because he uses the R=Cp-Cv rather than the correct nR=Cp-Cv
Sir, if a question is given change in temperature and pressure, ask to find work done. can i know this is under which condition? isothermal, adiabatic, isochoric or isobaric? or non of it?
@@wintergu8952 You need to know the details of the process path to find heat and work. U, H, P, V, and T are all state functions that only depend on the condition. But heat and work are both path functions. This is what enables us to build engines and refrigerators, to take advantage of the heat and work being path functions, by following closed paths on the state diagram, to trade these energies. For instance, 1L at 200 kPa expands to 2L at 100 kPa. In process 1, it follows an isothermal expansion and does 139 J of work. In process 2, it follows a linear path on the PV diagram, and does 150 J of work.
Hi again. In the process where P and V are changing, you write the internal energy (after Q=0) as PdV only and then equate this to n*Cv*dT. How come you safely neglect the VdP term? Is this because the system internal energy changes via PdV only when it does work? You say that you take infinitesmally small steps for V (dV) and do you assume that the pressure corresponding to these steps nearly does not change? (hence you write PdV). Thanks for your time.
Burc, Each thermodynamic process is unique and must be treated differently. In this video we have an adiabatic process which is defined as occurring so fast that there is no heat exchanged, therefore Q = 0. Since Q = 0, that means that delta U = 0 - W, which means that W = - delta U (which means that work done by the gas is equal to the negative amount of the internal energy change, which means that all of the energy used to do work comes from the internal energy of the gas) You can still find the work done by W = P dV, but then you have to integrate and you have to know the equation of the curve, which in this case is not known. That is why we find work by this equation: W = - delta U.
The change in internal energy of the gas is ALWAYS = n * Cv * Delta T regardless of the thermodynamic process. We ONLY use Cp for the heat added to the gas or heat removed from the gas during an ISOBARIC process.
I thought that work is the integral of P*dV, why do you integrate everything afterwards? Also in my book it is written as P*dV + V*dP=nR*dT can sb tell me why?
Your book is correct. Since both P and V are variables you must use the product rule. However you can also solve the problem as shown in the video by making a variable substitution for V in order to eliminate one of the variables.
You say that "if you multiply something with a logarithm, you can make that something the exponent". How can this be?? I mean, 3 * ln(5) is not the same as ln(5)^3.
I see that the error lies in the lack of parantheses. I agree that 3 * ln(5) = ln(5^3), but 3*ln(5) is not ln(5)^3. Due to video having no use of parantheses, I didn`t quite understand. Thanks for helping me clear that up.
Magnus Skeide Magnus, No parenthesis are required. Writing ln 5^2 is perfectly fine (at least in our texts it is). Check your math text book and see if that is the case for you as well. (If you could let me know I would appreciate it) Thanks
Benjamin, The change in the internal energy of a gas is always: delta U = n * Cv * delta T where Cv depends on the type of molecule in the gas: Cv = (3/2) R for a monatomic gas Cv = (5/2) R for a diatomic gas Cv = (7/2) R for a triatomic gas
Hi, i can not understan why is alway.....this should be only in a constan volume, but what happen if there is a constand external ptessure...i mean a cylinder in expansion or compression. In this case thw volume is not cpstand and despite of that we say. Au=ncvAT....why we cannot say AU=ncpAT?
Benjamin Diaz Benjamin, The internal energy of a gas cannot be dependent on how you get to that state. It can only depend on the temperature and what type of molecules the gas consists of. Start with the first law of thermodynamics. delta Q (heat added to the gas) does depend on how the heat is added. Delta Q = n Cv delta T if the volume remains constant and Delta Q = n Cp delta T if the pressure remains constant. I think you are confusing the two.
Benjamin Diaz What happens when you add 1 calorie to 1 gram of water? Delta U = m * c * delta T That is the same as adding heat to a gas except we use n instead of m and Cv instead of c (because with a gas it depends on the structure of the gas molecules.)
