First Law of Thermodynamics, Basic Introduction, Physics Problems

This physics video tutorial provides a basic introduction into the first law of thermodynamics which is associated with the law of conservation of energy. There are two main ways to change the internal energy of a system - that is, through heat and mechanical work. This thermodynamics video tutorial contains plenty examples and practice problems on heat, work, and internal energy calculations.
Open Vs Closed Vs Isolated System:
• Open System, Closed Sy...
Isobaric Process:
• Isobaric Process Therm...
Isochoric Process:
• Isochoric Process Ther...
Isothermal Process:
• Isothermal process The...
Internal Energy of an Ideal Gas:
• Internal Energy of an ...
_________________________
Adiabatic Process:
• Adiabatic Process - Wo...
PV Diagrams:
• PV Diagrams, How To Ca...
Thermodynamics Review:
• Thermodynamics, PV Dia...
2nd Law of Thermodynamics:
• Second Law of Thermody...
Heat Engines:
• Heat Engines, Thermal ...
Converting Heat Into Electricity:
• Thermodynamics - Conve...
________________________
Carnot Cycle:
• Carnot Cycle & Heat En...
Otto Cycle:
• Otto Cycle of Internal...
Refrigerators and Heat Pumps:
• Refrigerators, Heat Pu...
Entropy:
• Entropy Change For Mel...
Heat Engines and Refrigerators Review:
• Carnot Heat Engines, E...
Physics PDF Worksheets:
www.video-tutor.net/physics-b...

[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/

[@user-vf5zd5om3x]

Please I don't how to follow this link

[@strangeperson700]

When I get my degree it's going to look like I got it from the University of The Organic Chemistry Tutor! Lol

[@lavina3535]

Same for me lol!

[@donwalegere7211]

Thank you ,as a grade 11students taking science class I learn so much from you.

[@matrixate]

For the record, and as a note to myself, in example 1, the work done BY the gas should be looked upon as a decrease of internal energy used to cause the piston to move, in this case, up. So while you were probably taught that work done by a gas is negative, it depends what version of delU you're using. If it's U=Q+W, the work by the system is negative to represent that the system is consuming energy to cause motion in which case you plug in a negative quantity, whereas, in this example, the negative is already there and thus since work is Fd, and the direction of the motion of the piston is in the direction of the Force applied, then work is....positive. Right. So, work is positive in the equation that represents positive work done by a gas, hence U=Q-W. If this makes sense, great, but it does to me...rather, this is the way I see it.

[@bribrismathtutoring1909]

Makes a lot of sense. Thanks!

[@okaforcaritas7723]

U are really good… u have done two videos regarding this topic for physics and chemistry (view point from the surrounding and system) and u never contradicted urself… most textbooks contradict themselves

[@ma.patimarempis5338]

My teacher taught me when work is positive it is done by and if negative done on.

[@venkatavaibhavyejju2860]

What's wrong in it🤔😑

[@martinsclement8352]

That's what it is

[@divinechendo383]

@@NAfiqah99 wrong

[@divinechendo383]

@@NAfiqah99 what your saying is very wrong😂

[@leemonkumar9998]

Bro differences in subject

[@meera0021]

You r multi Skill person..(science, Physics,chemistry, Electrical, Mechanical) Great😎😎😎

[@leakry2164]

I like it very much. The first low of thermodynamics is good for me that I want to research and learn more about the fluid of mechanics.

[@owarida6241]

many thanks this really help me alot

[@jstroner]

Thank You. You are the smartest man on KZbin by far.

[@redbean9410]

i believe i am smarter cuz i got an A++ on my fifth grade math exam once

[@venkatavaibhavyejju2860]

@@redbean9410 🤭🤭😉😂😂

[@soretimekbib9637]

U ar my best teacher to all calculation need subjects i really thanks u my life saver

[@KeeraAnne]

1 month to my final examss and you helping me alot thanks 😊

[@terryivinho3228]

Great work BTW. but It would be nice if you put if the pressure was gauge or absolute in the last question .

[@hayatjemal3686]

You are smart man thank you 🙏

[@dilshadahamad593]

Fantastic your skills

[@user-rp3ur5er1v]

Thank you

[@johnpaulreyes7116]

Thanks Sir I learned many things so much from this video .Have a nice day 🎉❤🎉

[@NGOMA4LYF]

6:58 i think if its constant pressure then h=q-w then you can find u from h=u+pv

[@josphatporuzere4415]

thanks .. helpful

[@JohnRhynzMLasac-zq1si]

Thanks you I can now answer my Science

[@EggPuffsEdge]

Thanks bro nice voice

[@TheJProducti0ns]

I'm taking this class next semester (Electromagnetism and Thermodynamics). I want to study ahead during winter break. Could you point me to your videos that will help me for the first few weeks ?

[@SaifAMurad-qx8wd]

bro are you alive. how are you now

[@greatainoko1694]

How did it go

[@andrewjustin256]

7:15 Shouldn't work be negative because work is done by the system because it is expanding on heating?

[@Tamoor622498]

Exactly 2 years later, what a coincidence.

[@netsanettirusew4211]

My teacher taught me when work is positive it is done on and if negative done by

[@turkishandmore798]

I have some doubts though Work equals to the integral of Pressure as a function of Volume and not Pressure multiplied by Delta V in the first place , since dW equals to PdV

[@barrybobola3170]

yes but if pressure is held constant as stated in the video the integral simply comes out to pressure multiplied by the change in volume, so he is correct in making that statement

[@jodilyngarcia2199]

What is the illustration of number 2 and 4? please answer

[@akinapallydeepika4130]

Sir in 3rd question , they also gave " on the gas " , so then we get the value as -1350

[@maryb201]

Yeah. I just got 10/20 because I followed +1350 J. It should've been negative

[@a1aska466]

I think the work in the 3rd question should have a minus sign because gas expands so the system (the gas) is doing work which means energy is lost by the system to its surroundings in the form of work hence the minus. I might be wrong but you should check out that question again, just saying...

[@agustinjaysoncadag]

Correct

[@varshakalmanoor5416]

3 rd question why is the work done positive ?

[@hanissofia1922]

If i change the pressure to unit Pa why don't get the same answer?

[@kjpapa12]

why wouldnt you convert atm to pascal first in the last question. I did this and it made my answer much bigger than it should be idk what to do

[@zeki3661]

on the 3rd quesion why did u do only work dont on the system, it says on or by the system

[@kenuking369]

does anyone can help me? my answer in q3. in A. -1350J and in B. 2850J am I correct?

[@Armand__]

it's 11 pm and i have a quiz at 11 am tomorrow.

[@divinechendo383]

Please reply me What if your course is just "thermodynamics"? How will you be able to say when "W" is positive or negative

[@sihambelmiloud3351]

Like mine lol , We have it positif btw

[@nickoemmanuelvillacruz3090]

If you are Chemical engineering q+w If you are mechanical engineering q-w

[@agustinjaysoncadag]

Sir, as discussed in your previous video. In Chemistry, when work is done on the system it is positive and when work is done by the system it is negative. What happened? Edited: Okay, I get it.

[@gbemisolarahmon54]

Dat is y its different subject

[@retsepilemothepu1396]

My Physics improved from 50% on semester test 1 to 97% on semester test 2🎉🎉 Thank you sir, I have nothing to offer. I’ll thank you properly in the future.

[@agustinjaysoncadag]

As the system expands, when work done by a system it has a negative value. Shouldn't it be -1350J?

[@ahmeds4]

Why does it look so simple?!

[@adrianmoreira5603]

my teacher taught me the reversed sign convention, like work done by the gas is negative because it is losing energy etc. I think this is what the college board convention is in AP Physics II now

[@Franky10207]

steam table lesson please

[@tylerrussell7560]

I thought the sign conventions were the same. O.o

[@user-to3tr4tk1f]

Excuse me sir to the point of calculating the work l’m with u but as a chemist we apply the first law with positive charge delta U=q+w When l applied to the value of work, the value that l got it b not like yours Delta u=1500+1350=2850j that’s what I got 🥺 Can u explained?

[@kenuking369]

Bro, we got the same answer.. I hope someone will notice our questions :(

[@dinedd4866]

Can someone explain me the 9:17 why does 1L . atm is equal to 101.3 jules, where did it come from?

[@BIYJ_]

constant

[@mansichorwade7992]

In the second part why here why you took work done negative

[@tejdesaii]

Work done on the system is negative as we have to supply it.

[@wabidagim6219]

A lump of lead with mass 0.50kg is dropped from a height of 20m onto a hard surface. It doesn't rebound but remains there at reast for along period of time. What are? A) deltaQ,b)delta W and delta U for the leadduring this process?

[@AK-op4be]

OH, i didnt get what they meant with "work done BY a system is negative and work done ON a system is positive" but it's applied when doing a practice problem and it says for example "-if the system performs 200J of work-" .. you'd plug in W in the formula as a negative, because the word problem tells you that the system is doing the work 🤯

[@AK-op4be]

i gotta be explained things as if I was a toddler other wise I wouldn't understand 😵‍💫 but im proud I figured it out

[@Precious-ur1cl]

God bless you blud 😭❤️

[@majimbuuu1521]

How is W = 1350 in the 2nd problem somebody please.

[@omarahmed3621]

it is not he put -300? ??man

[@turkishandmore798]

applying this rule : Work equals to the integral of PdV where P is the pressure that’s constant in the case above , taking it out the integral leaves us with the integral of dV which is simply V that’s bounded between 0.02 and 0.05 . you get W=P x ( V2-V1) where P=45,000 Pa 45,000(0.05-0.02) gives out the value of the work which equals to 1350 J hope this helps 🖤

[@kylenason]

I'm so confused. I keep running into the first law being delta(u)= delta(Q)-delta(W). But I swear my class keeps teaching it as delta(E)=delta(u)+delta(Q)+delta(W) or even as delta(E)=delta(u)+delta(kE)+delta(PE). I don't even know what to think at this point. I'm just utterly confused.

[@usaball9193]

Just tell me when W will be equal to (-Pdv)?

[@nickoemmanuelvillacruz3090]

When pressure is constant

[@mhiztavicky1240]

Caleb said how??

[@jennylynosorio556]

Am still bothered by the signs for the “work done by the system and work done on the system”.

[@agustinjaysoncadag]

Same 😭

[@muhammadamirimranbinhazmir4113]

i think W=-ve (work done by system) while W=+ve (work done on system)

[@hamzashehata1632]

Other way round

[@caskie7465]

That's theoretically correct in chemistry since, the energy (in form of work) from the system if exhausted would be negative or subtracted, and vice versa. In physics it's different, because we deal with values that should be applied in the real world therefore the use of + for work done by system is valid and correct.

[@mesquite4052]

ΔU = Q − W or ΔU = Q + W ??, im confused...

[@imaginephoenix3283]

For physics, it follows the concept of Q-W, but if you insert the correct signs for Q and W, refer to 1:21 for these, you’ll be given the correct outcome. If you get confused about what sign the solution should be, you can remedy that by directly inserting the sign for Q and W into the calculator, if needed or by stating the signs on paper so you can take them into consideration while calculating. So it can be (-Q)-(+W) or (Q)-(-W), etc

[@turkishandmore798]

take it easy , the law can be written either as Delta U = Q+ Wexternal where Wexternal= -W ( which means that the work is being done by system , opposite to the case where Wex=W or positive so that the work is being done on the system ) or as Delta U = Q + (-W) same as delta U = Q-W . back to the point that Wexternal= -W Hope it helps 🖤

[@nickoemmanuelvillacruz3090]

Chemists thermodynamics is based on surroundings. Engineering thermodynamics is based on the system

[@strangealiengodofpower7628]

u r god

[@aldrinfrago9071]

7:26

[@frelandjaligue55]

Someday i will be a engineer

[@user-go6vi2xz9g]

❤❤❤❤

[@hanskz09]

anyone else heard the police siren in the background at 7:49.

[@ebhojayejuliet9728]

Yb better

[@ma.jeanabandoy8130]

What is this?? This is wrong men

[@Kurthustlet]

why?

[@aldrinfrago6015]

7:26