Physics 27 First Law of Thermodynamics (21 of 22) Summary of the 4 Thermodynamic Processes

[@byhariee]

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.

[@MichelvanBiezen]

Thank you. It is great to interact with our viewers.

[@Justaguy-d4n]

because he is the goat

[@PunmasterSTP]

How's school been going over the years?

[@collinthomas7901]

I’ve been stressing about my physics final all day, thank you for this

[@MichelvanBiezen]

You’re welcome 😊

[@PunmasterSTP]

How'd your physics final go?

[@PunmasterSTP]

This was a great recap, and I love how you kept the summary in the upper right part of the board intact over each of the videos!

[@MichelvanBiezen]

Thank you. Glad you liked it.

[@spoon4974]

Seriously, you're a real life hero for this video. College students everywhere are grateful.

[@MichelvanBiezen]

Not a hero. Just glad my wife and I can give back a little to the world.

[@emanueleg.4106]

COLLAGE???? it’s 10th grade here in Italy. 🙃

[@flydragon0804]

@emanueleg.4106 it is college if the high school student did not take the AP version and pass the exam for credits.

[@MichelvanBiezen]

Samual, The equation for delta U (change in internal energy) is always: delta U = n * Cv * delta T

[@haelrapal733]

sir, shall we use n or m?

[@winner1338]

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

[@anilsharma-ev2my]

What is Aneroid barometer efficiency ? Please found for shake of universe ?

[@mohamedmagdy299]

∆

[@winner1338]

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.

[@PunmasterSTP]

How'd your thermodynamics class end up going?

[@winner1338]

@ wow this is long time ago. Actually not really good. My advice to you is stick with the lecture of your professors. Otherwise you'll probably lose more time.

[@PunmasterSTP]

@winner1338 I'm sorry to hear it didn't go well. I took thermodynamics a long time ago and was just coming back to learn a bit of stuff mostly for my own gratification. FWIW I do think videos like this can help a lot of students.

[@billymushoke6370]

You Tube is really helpful to us because it makes us understand things even lectures and anything you would like to research about .

[@mostafaabdelrauof3185]

God bless you man , "your follower from Egypt"

[@nigelvanlaethem3860]

This is probably the most important video in this chapter. Thnx!

[@MichelvanBiezen]

Definitely a good summary. 🙂

[@del8548]

Proof that knowledge never dies. After 11 years this video still helps other people even myself. Thanks sir

[@MichelvanBiezen]

It's great to hear the video is still helpful to people!

[@MBhoodai]

Great summary! Very helpful to recap information a few days before my thermo I exam. Thank you.

[@tamashamas6193]

Cleared up a few confusions I’ve been having thank you so much

[@MichelvanBiezen]

Glad it helped!

[@AlmostCow]

Thank you for these amazing classes..

[@sandisomatu1131]

I've been stressing over my computer skills exam all day thank you for this

[@MichelvanBiezen]

Glad you found our videos. 🙂

[@ValiBee87]

This was extremely helpful , thank you so much

[@zainkhan-xz6mc]

A really amazing and helpful video to build up the concept. Sir can you make a video on the polytropic process.

[@Worldmemes4]

Another semester...and we are starting with thermodynamics...good work my eistein

[@MichelvanBiezen]

We have a lot of videos on thermodynamics. Enjoy!

[@Worldmemes4]

@@MichelvanBiezen Nice i will check out ✍️✍️✍️💪

[@abrahamabraham2583]

very simple and easy to understand thank you

[@gauravkotkar3731]

Excellent summary!

[@MichelvanBiezen]

Thank you. Glad you liked it! 🙂

[@likestomeasurestuff3554]

I appriciate this channel

[@pipertripp]

that tie is excellent!

[@latifurrahmanzihad]

love you,the great teacher............

[@maxtabmann6701]

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?

[@MichelvanBiezen]

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.

[@maxtabmann6701]

@@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!

[@MichelvanBiezen]

That is correct.

[@capcan23]

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

[@MichelvanBiezen]

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.

[@parisashahabi5067]

Thanks for the clarification!!

[@MichelvanBiezen]

Glad it helped.

[@mattbiss6414]

Great summary, thanks!!

[@yogeshmandrekar1777]

Very well explained…thank you

[@MichelvanBiezen]

You are welcome! Glad it was helpful.

[@SEAndies]

Thank you!

[@ganeshgs2241]

Superb sir!

[@MichelvanBiezen]

Thank you

[@stephanihab]

Thank you sir for your huge effort

[@sarujanrupan4831]

At @5:04 how come it's Csub(v) and not Csub(p) as pressure is constant?

[@MichelvanBiezen]

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.

[@sarujanrupan4831]

@@MichelvanBiezen Mm ok thank you.

[@fitrijaafar7091]

Thank you so much sir

[@AndreaNBAYER]

just excellent !. thanks for all the videos

[@JohnVKaravitis]

5:24 Delta U = n*Cv*delta T is ALWAYS true!

[@carultch]

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

[@Eric-sq4hd]

this guy is a true baller

[@YeshanewTeamru]

NICE TO SEE YOU

[@MichelvanBiezen]

Hope you like the videos. 🙂

[@isabellambert1011]

Love this summary, I just copied it down, now I’m gong to go take my quiz. Thanks.

[@bhavyaramakrishnan801]

In an isochoric process how do you know the change in pressure is constant and not exponential?

[@MichelvanBiezen]

That is so by definition. An isochoric process is defined as the volume changing while the pressure remains constant. W = P x delta V

[@quentinspoonmore4581]

​Typo...isochoric is constant volume while pressure changes. Isobaric is constant pressure while volume changes.

[@miracle99xD]

Isovolumetric process and isochoric process , both are same thing ???

[@MichelvanBiezen]

That is correct. Older text books use the term "isochoric", but newer textbooks use the term "isovolumetric"

[@vivekkumargupta1221]

Splendid ....

[@sandisomatu1131]

Where be the thermodynamics playlist

[@MichelvanBiezen]

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.

[@felino5334]

The man literally saved my physics final

[@MichelvanBiezen]

Glad to hear it. Good luck on your final! 🙂

[@qiqiz4290]

thank u so much! ur videos are very useful!!!

[@MichelvanBiezen]

Glad you like them!

[@chriskoperniak784]

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.

[@MichelvanBiezen]

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)

[@AkshayVasant]

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.

[@MichelvanBiezen]

It is done both ways, and varies by text and author.

[@sanft6070]

Great video, i love your bow tie!

[@MichelvanBiezen]

Thank you! 😊

[@lini4865]

This is great!

[@cristinamarquez3348]

Thank you sir.

[@MichelvanBiezen]

Most welcome

[@shinkizoku970]

Sir, how about polytropic process?

[@MichelvanBiezen]

What is the question?

[@clink8sl]

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

[@miracle99xD]

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..

[@MichelvanBiezen]

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

[@izzatiazan6494]

@@MichelvanBiezen so which one equation should we use +/- work done to substitute values and rationalise thermodynamic processes?im very confused cuz there are a lot of contradicting sources

[@izzatiazan6494]

@@MichelvanBiezen can u explain why u decided to use dU=dQ-dW?

[@almacyrenable]

Is the isovolumtric same as isochoric process

[@MichelvanBiezen]

Yes it is. The current editions of physics textbooks now use "isovolumetric".

[@favvee]

If work is done on the system, is it going to be negative sir

[@MichelvanBiezen]

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.

[@binyamin3747]

very good sir van

[@adarshvlogyzfr15v36]

well explain sir .. thank u

[@youssefrefaat1594]

Sir, please read this question. What are these red lines that you draw on the graph?

[@MichelvanBiezen]

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.

[@danmcginn9002]

Do you call the process of converting a solid to a liquid through a change of pressure 'melting', or is there a more precise term?

[@MichelvanBiezen]

It is technically a change of phase due to the change in pressure.

[@Misba14..7]

Sir how every PV graph gives workdone because p should be external but in these processes pressure given are internal

[@MichelvanBiezen]

A gas can do work by expanding.

[@laithNAli]

Thanks.. ❤️🇸🇾

[@fatmajjjxx2509]

So nice

[@iraqson666]

Thanks so much sir.. big greeting from iraq

[@linakasi403]

Thanks a lot

[@huzaifamehmood2966]

Hi could you tell me the relationship between the pressure and temperature in an adiabatic process please?

[@MichelvanBiezen]

TV^(gamma - 1) = constant

[@johannwegmann4365]

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.

[@carlosnunez-eddy2533]

what is the significance of the red lines on eitherb sides of the isthermic and adiabatic P x V graphs?

[@MichelvanBiezen]

The red line are isotherms (along which the temperature is constant). That way you can compare the isothermic and adiabatic process.

[@nellvincervantes3223]

On isothermic sir, Q = W, will it violate 2nd law of thermodynamics? Where all heat cannot be converted into work? Sorry im just confused.

[@MichelvanBiezen]

No it doesn't violate the 2nd law of thermodynamics. Only the heat added is converted to work.

[@pranaysai7713]

Sir, for isothermal process isn't delta Q 0 becoz delta t is 0???

[@MichelvanBiezen]

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.

[@pranaysai7713]

@@MichelvanBiezen sir that's ok but what about the formula?Q = nCdt?? Dt is 0... pls clarify

[@MichelvanBiezen]

That equation is ONLY valid for an isobaric process or an isovolumetric process, but not for an isothermic process.

[@ashw730]

Hi TY for this well explained video. Just a question Isnt delta U= W+Q My lecturer taught like that. 🤔

[@MichelvanBiezen]

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.

[@historyisthebest5831]

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.

[@sebastiandierks7919]

There is a typo in the thumbnail, should be a c_p (instead of c_v) for Q at the isobaric process.

[@MichelvanBiezen]

Good catch. Thank you.

[@peterv6i]

thank you

[@larryzdanis5377]

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 ??

[@MichelvanBiezen]

An isentropic process is just a special case of the adiabatic process. Therefore it would not represent a "fifth" process.

[@carultch]

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.

[@Andalfulfulde]

You're cool, What about ISOCHORIC processes😢

[@MichelvanBiezen]

Isochoric and isovolumetric is one and the same. = constant volume The term isochoric is not used anymore in most newer textbooks.

[@Andalfulfulde]

@@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

[@MichelvanBiezen]

Are you referring to the coefficients of expansion? You need to specify what the coefficients are for..

[@Andalfulfulde]

@@MichelvanBiezen yes sir

[@brave385]

where is polytropic process?

[@carultch]

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

[@iliekmathphysics]

why do we use cv even when volume is changing?

[@iliekmathphysics]

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

[@MichelvanBiezen]

The change in internal ALWAYS uses Cv.

[@Blessings-rw3qt]

Does anyone know why ∆U is sometimes written as equal to mC∆T and sometimes equal to nC∆T , which one is correct? Or when should either be used?

[@MichelvanBiezen]

When we are dealing witn gases it should be delta U = n c delta T n = number of moles For solids and liquids we typically use m for mass (in kg)

[@Blessings-rw3qt]

@@MichelvanBiezen alright, I think I get it now. Thank you.

[@farooquiphysics458]

Love from India 🇮🇳🙏 thanks sir

[@MichelvanBiezen]

Welcome to the channel!

[@wesleyblack8302]

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?

[@MichelvanBiezen]

No, that topic is covered at a later time. Typically we don't mention entropy yet, during the introduction to the adiabatic process.

[@ashleym9369]

for an isobaric process why can C_v be used for internal energy? I thought that was for constant volume only?

[@MichelvanBiezen]

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.

[@Peter_1986]

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.

[@MichelvanBiezen]

Laurelindo That is a good way of defining it.

[@historyisthebest5831]

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.

[@MichelvanBiezen]

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.

[@historyisthebest5831]

@@MichelvanBiezen Oh I get it! Thank you I understand now! Thank you!

[@historyisthebest5831]

@@MichelvanBiezen Yes but the question is saying that heat must transfer out in order to get a net work output?

[@MichelvanBiezen]

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

[@historyisthebest5831]

@@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!

[@briandiscala]

I'm curious, for an Isobaric process, wouldn't Q = nCv delta T and delta U = nCp delta T, as Cp is really Cv + R

[@MichelvanBiezen]

By definition, delta U is always equal to: delta U = n Cv delta T regardless of the process

[@elamaneceroscuro]

Bro but PV^5/3 is constant in adiabatic

[@MichelvanBiezen]

That statement was referring to the state variable. None of the state variables are constant in an adiabatic process.

[@bodotimesmusic4697]

Thanks a

[@Pad2Paper]

NOT ALL HEROES WEAR CAPES.

[@user-my8ll6lw4q]

my professor didnt need to everything sooo complicated

[@MichelvanBiezen]

Once you see that commonality and the reference to PV = nRT and the first law of thermodynamics it will make sense.

[@kartikey0208]

Reversible or irreversible process ? The most important point is missed lol😂

[@theprofessor2129]

Hehehe ideal gas is used in thermodynamics not reactive gases lol

[@abdelrahmanahmedfouad2084]

5 days left to my final physics exam

[@MichelvanBiezen]

All the best on your final!

[@abdelrahmanahmedfouad2084]

@@MichelvanBiezen thank you ❤️❤️

[@KashifKhan-xo7ym]

What is n sir

[@MichelvanBiezen]

n represents the numbers of mols of the gas

[@justin7even96]

why cant my prof just teach this????

[@PauloJAdams]

you are a g.

[@MichelvanBiezen]

Not in the least. Just a simple man with lots of faults.

[@rndungu1]

Making Soap

[@MichelvanBiezen]

Not sure we are following you?

[@theotichy9495]

Nice bowtie

[@MichelvanBiezen]

Thanks. 🙂

[@Thelydiariley]

Love this explanation, easy to follow

[@MichelvanBiezen]

Glad you liked it.