PV diagrams - part 1: Work and isobaric processes | Chemical Processes | MCAT | Khan Academy

[@Pr0x1mo]

Dude, you're up there with Khan in terms of how clear you are.

[@saratomorszki3904]

You clearly taught me everything in 10 minutes that my teacher couldn't in 45 mins

[@1polyron1]

Oh, I love these comments. They have me dying laughing because if I wasn't laughing I'd be crying at how hard it hits that what you said is facts

[@hashirkhattak9180]

Your voice is clear, your concept is clear ,,, you are perfection

[@1polyron1]

no you

[@dominicesteban3174]

As usual, David points out the pitfalls and traps like the Zen Jedi master teacher he is. Amazing!

[@hrperformance]

This guy is actually way better than Sal and Sal is a very good teacher

[@ariancruzat3519]

What a cool way of teaching. I am actually learning thermodynamics even though Im not taking it for now on my course. Haha. Thank you dude!

[@gingernut5057]

When I am older and work as a hobo I will thank you for this video

[@jaehoahn4810]

At 10:22, is it the work done should be positive? because the final volume is greater than the initial volume..

[@lalabEsjen]

You are really good!

[@amonahud3250]

Very clear explanation. Thanks a lot

[@khozikahmed2168]

amazing . fisrt time understanding those 4 processes

[@ShannTsaikf]

If it's an isobar compression, then shouldn't the work done be negative? And vice versa? Since for the isobar compression graph the change in vol would be negative.

[@foxeverett8164]

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[@drakecurtis8025]

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[@vihaanmenon7175]

No, because positive work is work being done on the system. So a compression is net positive, because you add energy to the system. While expansion is negative, since the system does work and thus loses internal energy. I think so, atleast. Learning this subject myself right now. Thought this might help someone else with the same question.

[@parkmin-su4793]

@@vihaanmenon7175 I'm sorry but aren't the conventions for work different in physics and chem? Like if work is done by the system=+W If work is done on the system= -W ? I'm confused about this part :")

[@carultch]

@@parkmin-su4793 Sign conventions are arbitrary, and set for the convenience of the context in which you are using them. You need to learn follow the sign convention given to you in your class. Ultimately, it's better to know how to state your answer with words, to explain the meaning of the sign you get, so that your answer is objective, regardless of what sign convention you are using. Not all physics and not all chemistry classes/books, are guaranteed to use the same sign convention as a standard. I'm accustomed to the heat engine sign convention, such that heat added to a system is positive, and work done by a system is positive, for thermodynamics problems. For the original definition of work, work done on an object is positive, and work done by an object is negative, so it is opposite.

[@Incrue]

Anyway thanks dude, i tough i understand this thing for the first time in my life

[@FeynmanAcademics]

Lolz. Me too

[@ECOMMUSK]

Thanks.

[@Incrue]

@11:30 you mean: because if its not equal the piston will move AND MAKE IT EQUAL AGAIN, right?

[@amaderjonnomathematics4288]

Sir i am not understanding anything what to do

[@rabiazafar5982]

Awesome....👌👌👌

[@haroldgaring8468]

please make TS diagram about these processes

[@Incrue]

@10:43 you make them be at equilibrium but just in the initial state, right?

[@leahevelee]

thank you!

[@zubaer73]

if the temperature in isobaric process increases then why the internal energy reduces? i found some maths refering the decrease of internal energy in isobaric process.

[@gtrslayer1]

Zubaer Ahmed because the gas inside the piston has to use energy to push all of that area upwards, like he explained in the video, a force is exerted through a distance.

[@gtrslayer1]

The examples of the isobaric processes you saw must have been ones where you remove some pressure from your piston and the gas inside pushed it up a distance without the need of heating it up. Thus decreasing the internal energy (U) of the system.

[@nthumara6288]

the world is runing on above 4 prosses approsimately

[@KA1blow]

06:12 if h changes, doesnt that mean the area does too

[@Mobimobipopo]

It's the area of the piston ( the lid, the circle part of the cylinder) not the area of the cylinder itself so it won't change when the h changes.

[@suryatejayalla3502]

Super

[@bleuemoone8710]

excellent!

[@merkov8715]

I need help. So what formula should I use? A=F/A(hA) or A=F/A(vf-vi)?

[@pulkitraju7034]

merkov ∆hA = Vf - Vi

[@joslyns6671]

Isn't work done on the system + and work done by the system -?

[@TMPChem]

Unfortunately this convention tends to vary among various scientific disciplines and even individual lecturers, so often it is necessary to carefully observe and note the sign convention being used in each particular case.

[@bjhu9031]

What software/app are you using for your presentation?

[@FeynmanAcademics]

When you know let me know too

[@itsasecret2132]

Delta Q = Delta U + Delta W right???

[@anon_ninja]

Sorry but if it moves to the left it’s positive work?? But isn’t change in volume negative? (Decrease in volume) and vice versa

[@arthikalexander316]

is it u=q+w or is it u=q-w?

[@AnirbanDas7]

+Arthik Alexander It can be both. Consider the expression: U= Q+W. 1. When work is done "ON" the system (e.g. External pressure applied to a gas in a piston-cylinder configuration to compress the gas), W= +W. Therefore, U= Q+ (+W)= Q+W. 2. When work is done "BY" the system ( e.g In a piston-cylinder configuration, when the gas exerts pressure on the piston, the piston moves upward), W= -W. Therefore, U= Q+ (-W)= Q-W. In short. Work done ON the system is always positive in nature (+W) and Work done BY the system is always negative (-W).

[@usamaqamar5562]

+Anirban Das Hey! does temperature increase in isobaric expansion?

[@AnirbanDas7]

Yes, it does.

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liked it

[@mshahzadaslam7357]

thats it

[@ParveenKumar-pg3ps]

nys

[@Sjsime827]

Are you God?

[@prashamranjan5638]

no

[@vortyx090]

Thanks.

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