Visit ilectureonline.com for more math and science lectures! In this video I will give a summery of isobaric, isovolumetric, isothermic, and adiabatic process.
Thankyou so much. This video was published 10 years ago when i was i like 4th grade and u still replying to all the doubts in the comments. Absolute Lengend.
I've watched a lot of vdo's about interference of light, including yours vdo's: Physics - Interference of Light The Thin Film. And i passed for physics. Thank you so much. Now i'm back for Thermodynamics.
this is the most asking question i will remember this, no matter what happens, sun explode, my cat dying, end of the world, my bike got stolen, no more pepsi, 1/0 is defined, etc. delta U will always be n * Cv * delta T
Mr Biezen, I just started thermodynamics, and your videos are really great help.I wonder if you can explain about Polytropic Process, I been taught that there are 5 processes, and polytropic process is one of them. Would be great help if you can. Thanks
Mubashir The general equation is: PV^n = C where C is a constant note that the ideal gas equation is: PV = nRT Draw the PV^n = C equation for different values of n and you'll see an interesting pattern. start with n = 0, then n = 1, n = 2, etc.
Thanks to Professor van Biezen, Please clarify in Isovolumetric Process,V=cte, what is de Heat Rejection in pulsejets,Lenoir cycle, Q, expression, regarding Thrust Force on 50 Hertz or cyles /sec.? Kind Considerations,Thankfully,Johann Wegmann.
why for p,v,t we associate the names with the change of the variables ex.(change in pressure/temp/vol = 0), but we just say heat = 0 instead of saying 0 change in entropy in the adiabatic case. is it to avoid students asking what entropy is?
At the 4:30 mark where you calculate the work done during an adiabatic process: how are you justified in stating that the volume remains constant (and hence the selection of your constant volume equation) ? I thought all 3 variables (volume, pressure, temp) are permitted to change during this process.
Chris I did not state that the volume stays the same. I used the equation for change in internal energy.(which uses Cv) (notice that V2 - V1 is not zero)
Do we use different sign conventions in physics thermodynamics? In my chemistry textbook, the equation is dU=dQ+dW In my physics textbook, the equation is dQ=dU+dW
Earlier negative sign was assigned by the work is done on the system and positive sign when the work is done by the system . this is still followed in physics books , although IUPAC has recommended the use of new sign convention..
dU = dQ + dW: The change in internal energy of the gas is equal to the heat added to gas + the work done ON the gas. dU = dQ - dW: The change in internal energy of the gas is equal to the heat added to the gas - the work done BY the gas
The red lines are called: "ISOTHERMS" They represent lines on the PV diagram along which the temperature remains constant. An isothermic process is drawn parallel to isotherms.
This is a fantastic comparison of these processes. Please verify where an "isentropic" falls amongst these.. I know that isentropic is both adiabatic and reversible, so would it be the same as the adiabatic -does the adiabatic equation for work presume reversibility ??
Yes, the adiabatic equation for work assumes reversibility. Adiabatic and reversible, collectively form the term isentropic. If there were irreversibility, it would mean there are inefficiencies in the change of internal energy and work done. The extreme case of an irreversible adiabatic processes, is called an isenthalpic process. The Joule-Thompson effect is an example of this.
It is Q + W when it is work done ON the gas and Q - W when it is work done BY the gas. The first law of thermodynamics is expressed both ways. I prefer the second way.
Yes so when work is done by the gas, the W is positive and the equation is delta U = Q - W. The minus sign is not representing the sign of W so W is still positive here. But if work is done ON the gas, then W becomes -W. delta U = Q - (-W) equals delta U = Q + W.
In Isothermal Process Change In temperature Is ZerO B/c Of Constant as We Know Q=MC Change In Temperature So We Say temperature Is Constant For Isothermal Process Why Heat Become Zero? pls Teacher Answer It It Confused Me A Lot
In an isothermal process, the temperature remains constant, which means that the internal energy of the gas remains the same, which means that if the gas does work, (by expanding), then this energy needed to do work must come from somewhere else (head added to the gas). Therefore W = Q (the work done by the gas is equal to the heat added to the gas).
Hi Sir, since you are teaching thermodynamics from the pov of Physics, I think it will be better if you describe work as "work done by the system" instead of 'by gas', because I got confused for a moment, if this was chemistry or physics. I hope I made sense and sorry if I didn't.
Yes, that is a common source of confusion. Cp only applies to the heat exchanged (Q = n Cp delta T) But regardles of the thermodymic process we always use Cv for delta U Delta U = n Cv delta T for all processes.
Are you able to tell me how you practically change the volume without changing the pressure? My professor always avoided this detail. And later I found no book that looks at this question. My thoughts to this are all useless. I thought one could punch a hole in the back side so that the exess air can escape, but then this would mean that the pressure outside must be the same as inside, otherwise the volume would adapt to the outside pressure. But if the pressure outside is the same as inside, whats the point of defining a volume here?
Take a container and place a sliding lid at the top (like the piston of a car cylinder). Place a mass on top of the lid, such that the force remains constant and hence the pressure remains constant. If you heat the gas, it will expand under constant pressure.
@@MichelvanBiezen I see! Thus you do not change the volume yourself, but allow it to change when it gets heated or cooled, using a piston that excerts a constant force. Thanks, this makes it all understandable. The temperature does not change due to a change of volume but it gets changed by adding thermal energy. Thus cause and effect are reversed. Cause is the heating and effect is the volume change. Thanks!
Ashley, Yes, this is often confusing to students. The internal energy of a gas is only dependent on the temperature and Cv. The internal energy of a gas should not depend on how the gas got into that state. But the heat added or taken out of a gas does the depend how the gas changes (via constant volume or constant pressure). So remember, delta U = n Cv delta T.
Michel van Biezen I have always thought of it like this - constant volume means that no work is done, therefore dW = 0, and dU = dQ. And since in this case dQ = n*Cv*dT, this also means that dU = n*Cv*dT, and thus dU = n*Cv*dT would then be true for all other processes as well. Something like that.
It depends on how "it" is defined. First we need to know what the "it" is. Let's define the first law of thermodynamics: The change in the internal enegy of the gas is equal to the heat added to the gas minus the work done by the gas. Using that definition, when the gas does work it is positive work and when work is done on the gas (by compressing it) it is negative work.
There are about 10-15 thermodynamics playlists on this channel. They can all be found easily from the home page. Let us know if you have trouble finding them.
Hi sir sorry just a question. In my textbook, there's also a statement says that in every cyclical process, heat transfer must occur from the system in order to get a net work output. But why is that? Been digging Thermodynamics lately and sometimes my brain is just slow.
Since the internal energy of the gas is the same when you end up at the state you started from, the energy needed to get the work done must come from somewhere else. (therefore the heat gained) First law of thermodynamics: change in internal energy (which = 0 for a cyclic process) = heat added - work done No heat added means no work done.
In order to perform work, heat must travel into and out of the process. But more heat goes in than comes out. The difference is the work done. I suggest you watch the rest of the videos and this playlist: PHYSICS 27 FIRST LAW OF THERMODYNAMICS and PHYSICS 28 CYCLIC PROCESSES and PHYSICS 29 EFFICIENCY OF HEAT ENGINES for a good understanding of this topic
@@MichelvanBiezen Okay thank you!! Actually I already watched almost all of the videos in your suggested playlists but I will watch them again. Thank you though!
Polytropic processes are the general case where P*V^n = constant. n has nothing to do with the number of moles, but is called the polytropic index. It can be any number from 0 to +infinity. All of these processes are special cases of the polytropic process. Isobaric, n=0 Isothermal, n=1 Adiabatic/isentropic, n = k, where k is a substance-specific constant. For air and 2-atom elemental gasses, k=1.4. Isochoric, n=infinity
That is an approximation that Cv is independent of temperature, which is approximately true, but not completely true. The complete formula is delta U = n*integral Cv dT, from T=Ti to T=Tf. Or: dU/dT = n*Cv
okay here’s what I’m thinking: (I’m not taking thermodynamics at the moment). I think it is that “delta U = Cv delta T” is simply the definition of Cv and therefore it can be applied to any process since U is a state function. Also the ‘constant volume’ aspect of the specific heat capacity only depends on the fact that we’re dealing with gases. Assuming that these are ideal gas conditions, Cv would be the same for monatomic gases, and same for diatomic gases (though different for each category) since the only type of energy contained in it is kinetic energy, which is directly related to temperature and internal energy
In an isothermal process the internal energy remains the same (delta U = 0) since the temperature is constant. But when the gas expands, it is doing work. Since the energy cannot come from the internal energy, it must come from the heat added to the gas. Therefore work done = Q received.
@@MichelvanBiezen okay Engineer, I appreciate you. Please what about "betta and kappa" coefficients for university thermodynamics, am searching for it on your playlist but seeing nothing
