PV diagrams - part 2: Isothermal, isometric, adiabatic processes | MCAT

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PV diagrams - part 2: Isothermal, isometric, adiabatic processes | MCAT | Khan Academy

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Пікірлер: 104

@secondo92 7 жыл бұрын

Holy sh** this was so well explained that even my cat would understand it!

@Destroyer9623 9 жыл бұрын

would be good to add that the reason an adiabatic process is steeper than the iso-thermal process is due to the decrease or increase in gas particle velocity (which affects pressure) caused by the decrease or increase in temperature.

@JhaAtiksh 4 жыл бұрын

Was literally gonna ask this, my g

@DaringToast Жыл бұрын

thanks for the info, i had just begun wondering why pressure ended up higher for adiabatic expansion

@AgEyal 7 жыл бұрын

You saved my life.

@nandhakumars680 5 жыл бұрын

Nice video which I understand this PV diagram for the first time thanks a lot good one 👍😍

@briannguyen6994 3 жыл бұрын

I wish there were more physics derivation videos from you specifically

@coolmonkey619 3 жыл бұрын

Why does this not have more views

@diegoarana5862 3 жыл бұрын

this is very helpful when your teachers method of teaching is reading word for word off a PowerPoint and jarringly bouncing between slides constantly.

@yazadpardiwala687 3 жыл бұрын

7:40 You are a) extremely good at your explanations and b)very enthusiastic/funny at times

@user-nk1cz4fk1p 8 жыл бұрын

Hi, please take note that there is a mistake at 12:11. both isothermal and adiabatic curves should be single arrow. during compression process, the work done of isothermal process is less than the work done of adiabatic process, so the adiabatic curve should above isothermal curve in the revers way.

@connecteddthoughts 7 жыл бұрын

No, that's not a mistake. He used two arrows to symbolize compression and expansion in both isothermal and adiabatic processes. At least, that's what I think.

@dragowolfx 9 жыл бұрын

ihave my thermo exam today thanks

@golddigger0325 7 жыл бұрын

Thank you for explaining so perfectly!

@cathytaaf4840 4 жыл бұрын

I am so grateful for finding this video haha. Thanks!

@ngumamah4477 2 жыл бұрын

So easy to understand

@abhinavrauthan5181 4 жыл бұрын

What a tremendous !!! video best explanation ever!!!😇

@DaveShree 6 ай бұрын

Do you think that internal energy remain same and work done by heat exchange in isothermal process, then think again. Heat exchange takes place when there is difference in temperature but in isothermal, temperature is same. Work done is at cost of internal energy, change in pressure.

@mikesmate6672 6 жыл бұрын

Hi quick version of getting work in an Isothermal process when given the change in V ,number of moles and the temperature is using the equation W=nRTln[vf/vi];

@sonalirajput832 4 жыл бұрын

Omg..... U guys are so good at teaching 👍👍👍👍🙏🙏🙏🙏

@makahragayle8973 6 жыл бұрын

Thanks for explaining these so well!

@rajendrapathak760 7 жыл бұрын

All other vids kept confusing me.TYSM

@hyuckkcuyh 2 жыл бұрын

soooo useful, I love this channel

@amirthavarshiniy6666 Жыл бұрын

I can't believe these videos are uploaded 7 years ago

@atulprakash416 2 жыл бұрын

excellent

@DuuuuuudeStopthat 6 жыл бұрын

I shouldve watched your vid before I spent 3 hrs of my life trying to figure out what these three are.

@chakia4347 3 жыл бұрын

Thank u

@vortyx090 8 жыл бұрын

Thank you very much...!!!!!

@clayton8945 4 жыл бұрын

This is a really good video. Very clear and easy to understand.

@user-rw5pb2zb3l 6 жыл бұрын

Do you now somthing ...u are awesome....thank sooo much

@RalishGaming 9 жыл бұрын

Hey! Just wanted to point out something. During your isothermal description you rearranged the formula wrong, should be U=Q-W. Great vid though! Helped out alot :)

@nurulshafiqah1 8 жыл бұрын

ohh thanks..im confuse for a little bit because of that while watching this

@kelvintan4674 8 жыл бұрын

U=Q+W is true because he said W= work done "on" gas... For U=Q-W, W= work done "by" the gas... its just a matter of how u want W to be represented by...😉

@cutler610 8 жыл бұрын

Awesome video, just what I was looking for!!

@aqsamustafabutt5400 7 жыл бұрын

rest of theexplanation was very good

@Starfighter-x2s 3 жыл бұрын

8:20 you mean deltaQ=U+W

@arstudios6883 3 жыл бұрын

V good

@lodi8294 4 жыл бұрын

thank you

@darkop8293 2 жыл бұрын

This was a good video but I think the equation for (delta U=Q+W) is wrong I think it's suppose to be minus

@bbmeme2010 7 жыл бұрын

Well Done - thank you for explaining thermo and making it easy to apprehend. Where were you when I was in undergrad?!

@hisyamhasbi8257 4 жыл бұрын

Is apprehend means the same thing as i think it means?

@cmhardin37 4 жыл бұрын

Thank you good sir. 👍🤘

@14aravindlochaner24 6 жыл бұрын

Thank you thank you thank you very much

@JuAnKing24 6 жыл бұрын

isn't it delta U = -W since work is done onto the system cause work isn't done BY the system?

@nayanikakundu7824 6 жыл бұрын

Actually, this depends completely on your perception. Conventionally, when work is done ON the system, it is taken to be positive, because, the system gains energy as, U=Q+W. On the other hand, when work is done BY the system, it is taken to be negative, as energy is lost by the system, giving rise to the relationship, U= Q- W. However, if you take the first law of thermodynamics as: U= Q-W, then, your assumptions is correct. If work done on the system is taken as negative, then, U=Q-(-W) i.e. U= Q+W. This completely depends on your perception, and honestly, convention does not matter as long as your conception is sound.

@barahui9 3 жыл бұрын

Thank you for this video. My professor thought it was cool to use a shitty graph and an unintuitive explanation to help us understand this.

@ravilkhateek4999 6 жыл бұрын

Well done ... Good explanation

@sarakhan3151 7 жыл бұрын

thank you!

@aqsamustafabutt5400 7 жыл бұрын

why adiabatic graph is more steeper?????????

@rickandelon9374 5 жыл бұрын

It is because of the instant action required to do it. NOT SURE BUT!

@carultch Жыл бұрын

Here's the derivation of the adiabatic condition: hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiabc.html#c1

@viraldunia7905 3 жыл бұрын

brother. what is the name of this Software. you are writing like pen by mouse Amazing.

@malawby 5 жыл бұрын

Does the GHE effect earth's temperature, or is earth's mean temperature of 15 C controlled by the atmosphere's mass and pressure from gravity?

@dipakmudi3877 6 жыл бұрын

thank u for help me

@geonwoo1213 5 жыл бұрын

nice!

@zaidahmed4800 4 жыл бұрын

helpfull

@MuhammadQasim-fu4hb 2 жыл бұрын

I thought the first law of thermodynamics was delta U = Q - W

@DonSyndrome 5 жыл бұрын

Damn good explanation. I studied engineering at Cornell. Trying to coach my son’s Science Olympiad Thermodynamics team.

@Abdulrahman-wo2ob 8 жыл бұрын

For some reason, I thought the isothermal & adiabatic curves are switched, meaning tha adiabatic requires more work. I guess I was wrong.

@muhammadhassamkhan2331 4 жыл бұрын

I am stuck with first law of thermodynamics because of its mathematical expression, some are using like above expression which is u= Q+w and remaining are using U= Q- W

@francisjosephlarita2818 4 жыл бұрын

They put negative on it because of W=-P(Vf-Vi) iguess?.

@suhani715 Жыл бұрын

It's just some difference we have in subjects ...like the plus one is for phy and negative for Chem.....it's just differs on tge respective of where you are seeing or basically the point of view :)

@nikko-kun643 6 жыл бұрын

wll explained

@123lebornjames 9 жыл бұрын

ohhhhhhhhh my god thank u!

@thushara765 6 ай бұрын

❤

@md.ahanafibnayakubsajim9782 6 жыл бұрын

This is a isothermal process. So the eqution should be like dQ=dW(where du=0)

@harlymoon452 6 жыл бұрын

Sajim Ahamed no I think it should be W=-∆U where (Q=0)

@md.ahanafibnayakubsajim9782 6 жыл бұрын

harly moon i need sometime to think about it.

@williamnathanael412 6 жыл бұрын

harly moon isothermal means there is no change in temperature, hence no change in internal energy, U. dU is therefore 0, yet heat might still flow, so dQ isn't zero.

@Akbarwayse 8 жыл бұрын

At 8:22, did you mean that U = Q - W?

@dingleberrys232 6 жыл бұрын

Although i didnt ask the question, thank you for your reply. I was getting proper confused, but now i understand.

@jairabelleventurillo9998 4 жыл бұрын

isn't it that the formula for 1st law of thermo dynamics is Q= change in U + WORK? how come that change in U is now equal to heat plus work. ? it should be Q-W ? please correct me if I'm mistaken. Thank you.

@Gudi923 4 жыл бұрын

The equation is U= Q + W and the W can be positive or negative depending on what the gas does. If the gas did the work, then work is negative. If the work was done on the gas then the Work is positive. Hope that helps

@hisyamhasbi8257 4 жыл бұрын

Holy fuck, thanks man, truly from the ravine of my heart!!!!

@astha_yadav 2 жыл бұрын

Holy Jesus if only i searched and watched these in 11th grade 😭😭 Edit: wasted money on shi*ty tutions

@engineerstrange4447 8 жыл бұрын

Hey can u please tell 🙏🙏🙏. Where do u write this I mean what is the software.

@arunapothula5577 4 жыл бұрын

water temp outside container increase or not

@iksiX99 4 жыл бұрын

yes, but very little, unless you have a small amount of it

@anirudhstarc7234 7 жыл бұрын

Thanks a lot!

@Arsalankhan-mv7ng 7 жыл бұрын

👌👌👌

@abhijitsingh07 6 жыл бұрын

But first law of thermodynamic States that dQ=W+dU , so how dU=W in an adiabatic process

@dingleberrys232 6 жыл бұрын

Do you mean why is it not dU=-W? Because I do not understand why it is not a negative W value.

@leftenanalim 5 жыл бұрын

Because in an adiabatic process, heat change is 0, hence dq=0. Therefore 0=dU+W W=-dU

@nithin884 7 жыл бұрын

How is pressure constant in Isochoric process?

@carultch Жыл бұрын

It isn't. It is volume that is constant in an isochoric process.

@hussienjaafar915 7 жыл бұрын

but when we compressed the tank W is negative, the equation is delta U= Q-W ( W is in negative since work is done "on system") thus it becomes positive and not delta U= Q+W and since heat is flowing out so Q is negative , just wanted to clarify on delta U= Q-W , please reply

@hh61 7 жыл бұрын

Why is that at a constant volume no work can be done but at a constant pressure work can be done?

@ChrisCardozaisawesome 7 жыл бұрын

Husam because at constant volume, change in volume is 0. the p🔺v term goes to 0 leaving u=q . so at constant volume, no matter what work you do, it wont affect your system.

@hh61 7 жыл бұрын

Ohhhhhhh, for the life of me I was super confused about it, but yeah if the final and initial volume stays the same then that will be 0. Now it sounds so simple!

@dheerajkumar-dq6qi 6 жыл бұрын

why adiabatic curve is more steeper than isothermal curve ???

@carultch Жыл бұрын

Good question. The adiabatic curve is given by the relationship of P*V^k = constant, where k is the ratio of specific heat capacities, defined as k=cp/cv. Since the constant pressure specific heat is always greater than the constant volume specific heat, due to the gas expanding as it is heated under constant pressure, the value of k will always be greater than 1. Isothermal is given by P*V = constant (straight from the ideal gas law), while P*V^k = constant generates a steeper curve that governs the adiabatic reversible process. Why does P*V^k govern adiabatic processes? Here's the derivation. hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiabc.html#c1

@charliedajoob2246 6 жыл бұрын

LOOKIT. NOW LOOKIT. LOOKIT

@FieldsSam1 9 жыл бұрын

Is the Q component heat transfer?

@billyroberts4316 8 жыл бұрын

+The man himself yes, thermal energy

@secreswallowtail3372 6 жыл бұрын

what about gamma ratio?

@carultch Жыл бұрын

The gamma ratio (some books call it k), is the ratio of specific heat capacities, that governs the reversible adiabatic process for an ideal gas. It is defined as gamma = cp/cv, as the ratio of constant pressure specific heat capacity, to constant volume specific heat capacity. It generally is a function of the shape of the gas molecule, such that 2-atom gasses like air and its components have gamma = 1.4, and single atom gasses like Argon, have gamma = 1.67. It has to do with the degrees of freedom in the gas molecule, that act as modes of energy storage.