Physics 32 Kinetic Theory of a Gas (1 of 10) Pressure and the Kinetic Model of an Ideal Gas

[@MichelvanBiezen]

The two books that I like to use are: University Physics by Young, Freedman, Sears, and Zemansky. The other is Physics for Scientists and Engineers by Serway and Jewett. There are many other good texts, but these are some of the better texts.

[@lt.ifeplumz1315]

These are the texts my institution recommended for us year 1 students.

[@TheAInfinity]

You sir deserve so much more view than this. You're pure genius in explaining physics concepts.

[@MichelvanBiezen]

Mohammed, Those are good questions. Off course for a few atoms the velocity in the x, y, and z directions are going to be different, but for trillions of atoms, it is OK to say that the average velocities in the 3 directions are going to be equal. The average impulse per unit time that the molecules impart on one wall can be found by dividing the number of collisions against the wall by the total time. That means the impulse per unit time is caused by 1 collision divided by the time the molecule takes to travel across the box (back and forth).

[@phi4321]

Your channel is highly underrated. Wish you the best in the future

[@MichelvanBiezen]

Thank you. We appreciate your comment. 🙂

[@MichelvanBiezen]

Most of physics is there, but I still need to add some missing sections in mechanics and waves. We are working on it

[@Libservative79]

I have a physics III exam tomorrow and this REALLY helped me nail down the construction of these equations and relations. THANK YOU!!!

[@PunmasterSTP]

I know it's been over a decade, but if you can remember, how'd your exam go?

[@lanmisu]

Thanks , very detailed explaination. Now I understand where that formula derived from. You save me a lot. Thanks a bunch. Hope you are healthy, wealthy and can upload more video with detailed explaination as always.

[@noehoungbe6355]

The best explanation of this formula we can have !! ❤

[@MichelvanBiezen]

Glad you think so!

[@agathernakiweewa3934]

I proved the ke. And the lecturer wondered. Surely I explained all the details on the paper. Thank u so much.

[@MichelvanBiezen]

Great 👍

[@mohamedsawan9354]

I just have two comments on the video The 1st is why vx=vy=vz The 2nd is delta t (in the equation of the impulse )which is equal to the time collision how can equal the time of the molecule's travel across the cube ? I hope quick response,please.

[@kaloomuwana9077]

I can deal with most of the assumptions fine, im just struggling with why we use the time it takes for the ball to travell from one side to the other as back, almost as if its the time of collision in the impulse equation

[@fairytal.s]

best explanation ever.

[@radhikaganapathy2716]

Awsome Awsome!! please start online tuitions for students overseas your bank balance will go bonkers!!!!

[@vagmahesh]

Just wanted to point out something that i rarely see being discussed in any lectures or books while they do this derivation. When was learning this what bugged me when we calculate force, we use delta t = time between successive collision? :O.. that made no sense to me.. clearly from mechanics i understood that, delta t should be the time OF COLLISION.. time it takes for that tiny winnie molecule to bump the wall and then reverse its direction.. its the TIME OF CONTACT.. and no one explained why.. Only later did i realize that.. what we are interested here is in the AVERAGE force .. so although we are right, that infact delta t SHOULD be the time of contact.. we see that, for a long time there is NO force imparted. so when i AVERAGE it out over THAT ENTIRE TIME it takes for the molecule to bump back.. we should take the entire time between collision :D.. its very hard to explain.. maybe that's why people leave it out.. but its very fundamental.. and as a teacher i hate to leave things like that.. anyways nice explanation kudos :)!

[@MichelvanBiezen]

Mahesh, Very nice commentary!

[@vagmahesh]

Michel van Biezen Thank you.. I am glad, i made sense :)

[@vagmahesh]

***** What more? :P

[@sumanthaluri8398]

Thank you, I came to KZbin with that exact question bugging me

[@pakheitsoi4886]

Thank you so much. I struggled with the same question in my mind until I saw your comment which enlightened me! Thanks a lot!

[@recall660]

I will take the chance and ask another question ,is those videos cover all the physics up to university level. For engineering. I mean is this all the physics topics ,thanks prof

[@guacamole3109]

Dear Michel, Great video, as usual! I have a question though... When you convert the equation from the force of one molecule to the force of all of them, you divide by three and then multiply by the number of molecules.... Please could clarify if you are doing this because; The combined velocity of v(x) v(y) and v(z) is 3V and hence you must divide it by 3 to find the average v of each direction, and you are multiplying by N to therefore find the correct total V. If not could you please explain the correct reasoning? Thank You so much!

[@MichelvanBiezen]

Yes, you need to find the average velocity in each direction.

[@guacamole3109]

Ah, thank you for your quick response :D

[@amansinghal2431]

sir I have a doubt..... when you wrote force equals rate of change of momentum in my opinion delta t should be only the time of the collision. and not the whole time that it has traveled.

[@MichelvanBiezen]

Yes, that is a difficult concept to grasp. I must admit, when I saw that for the first time, it gave me doubts as well. The best way to think about it is this: The momentum of a molecule is changed on the left wall once each time it makes the round trip. The impulse is the area under the force-time graph and that area will be the same regardless how much time you want to consider. (Only the time of collision which would make the force much larger, or the time to make the round trip which would give you the average force over that time, which is the actual force experienced on average).

[@amansinghal2431]

I think I understood 90℅ of the concept. thank you for the explanation and time you gave.

[@rjaph842]

@@MichelvanBiezen that explanation can suffice, thanks sir

[@rajusharma-co1cg]

Awesome sir.... Helped me a lot

[@uuitgaurav]

Today i watched your one, more video.Sir.

[@darka3274]

i have question. isn't n=m/M(mol mass) , and n=N/Na then how m*Na= M(mol mass) at 7:50 .

[@MichelvanBiezen]

m is the mass of a single atom (or molecule). Na is Avogadro's number. m x Na gives you the mass of a mol of molecules and is therefore called the molar mass. The number of moles n = N/Na but not the first equation you showed.

[@darka3274]

thanks sir

[@Uyry-m8m]

I wonder why time interval of Impulse equal to time interval of molecule moving back and forth? I think time of impulse is short and time of molecule moving is long?

[@MichelvanBiezen]

The impulse will occur once every time the molecule travels across the box and back. Impulse can also be calculated as change in momentum.

[@Uyry-m8m]

@@MichelvanBiezen sorry teacher, I am student. Could you detail? I am not sure and hard to visualize.

[@dishangdoshi9966]

HELLO SIR can I say that if the container is at rest the sum of {DEL .p(WALL)} net change in momentum of the walls{ left and the Right} is zero. Similarly For Y and Z axis. If I take three walls of the container THE RIGHT ONE( THE POSITIVE X DIRECTION WALL) THE UPPER FACE OF THE CONTAINER(THE POSITIVE Y DIRECTION WALL) AND THE FRONT WALL( THE POSITIVE Z DIRECTION FACING OUTWARD), IF THE GAS IS APPLYING EQUAL PRESSURE ON EACH WALL AND IF FORCE EXPERCIENCED BY THE WALLS ARE SAME AS IN STEADY STATE IF I EQUATE THEM FX=FY=FZ IE. 2M(VX)=2M(VY)=2M(VZ) (PER SECOND HENCE t= 1SEC) THEN CAN I SAY VX=VY=VZ?. BECAUSE IN STEADY STATE MOMENTUM IN EACH DIRECTION HAS TO REMAIN SAME RESULTING NET MOMENTUM OF ALL DIRECTIONS IS EQUAL TO ZERO OTHERWISE THE CONTAINER WILL MOVE... SINCE IN STEADY STATE SUMMATION OF VELOCITY SQUARE IN X= SUMMATION OF VELOCITY SQUARE IN Y= SUMMATION OF VELOCITY SQUARE IN Z.... AND SINCE IF I ADD THE NET CHANGE IN MOMENTUM OF EACH WALL, NET CHANGE IN MOMENTUMS OF WALLS IN X + NET CHANE IN MOMENTUM OF WALLS IN Y + NET CHANGE IN MOMENTUM IN Z IE. 2MVX + (-2MVX) + 2MVY + (-2MVY) + 2MVZ + (-2MVZ)= 0. NET MOMENTUM HAS TO BE ZERO SINCE THE CONTAINER IS AT REST....... IF THERE IS NO PRESSURE DIFFERENCE INSIDE THE MOMENTUM IN EACH DIRECTION IS SAME SINCE VELOCITY DISTRIBUTION INSIDE THE CONTAINER HAS TO BE EQUAL THE FORCE ON EACH WALL HAS TO BE SAME..... OR SUUMATION VX^2= SUMMATION VY^2= SUM,ATION VZ^2 ( CONDITION FOR STEADY STATE) THEN TAKING TWO EQUALITIES AT A TIME VX=VY AND VX=-VY VX=VZ AND VX=-VZ VZ=VY AND VZ=-VY HENCE VELOCITY IN EACH DIRECTION IS EQUAL RESULTING NET VELOCITIES IN ALL DIRECTION IS ZERO SIR CAN U TELL ME WHICH ONE OF THIS IS CORRECT FOR VX=VY=VZ

[@MichelvanBiezen]

Not sure what you mean by the "net" velocity. How would you define that. The average velocity and the root mean square velocity certainly are not zero. (If they were, the gas would not be able to exert pressure on the walls).

[@dishangdoshi9966]

@@MichelvanBiezen defining net as if molecules moving In right and some molecules moving in left Vx taken positive in right and Vx taken negative in Left . The overall net velocity of molecules In x direction is zero

[@dishangdoshi9966]

@@MichelvanBiezen I am saying by taking The wall and the container to be system

[@dishangdoshi9966]

@@MichelvanBiezen Yes sir Avg won't be zero my explanation was the force experience by the walls of container due to collision will be equal. So basically if the molecule colliding the wall creates a force on of the surface , the same molecule colliding the wall on the other surface ( Two extreme surface of the container suppose taken in x direction only) will exert a force on the other wall resulting The Net force on the walls of container in X cancelling out each other.

[@MichelvanBiezen]

The force on the left side of the container will not cancel out the force on the right side of the container. If you draw a free body diagram around a section of the left side of the container and determine all the forces on it, you will see that the forces on the right side of the container cannot affect those on the left side.

[@jennyispoop4]

Wow, the derivation of pressure's equation seems so simple now.

[@wjgoh5161]

Hi sir, just want to ask why mass times avogadro constant will become molar mass

[@MichelvanBiezen]

Mass of 1 atom multiplied with the number of atoms in a mol is the molar mass

[@ahmedal-ebrashy3691]

Mr and Mrs Biezen, ypu are my are my heroes. Thank you for making the World intelligent. There one thing you dont know about yourself but in speed 2x you are still very understandable. Some otheronline tutors lose that understandabilitybafter 1,75x That is you have super manly voice.

[@MichelvanBiezen]

Ahmed, that is actually funny. We didn't know that is a way to measure clarity. We'll try it.

[@alirezarahimi7956]

Thank you so much.🙏🙏🙏

[@MichelvanBiezen]

You found some more thermodynamic videos of interest.

[@nilufargh]

Firstly, Thanks you so much for your great videos! I've watched all of the videos for this chapter and I really enjoyed them. I just have one question, the way that you derived Pressure formula is a bit different than my Prof: His Way: deltaP= 2mc then in order to get F= 2mc(c/L) , so because c is only for one molecule he said the average is c^2= mean square velocity so now c= Nc^2. At the end in order to get P= F/ A so he said A= 6L^2 ( the Area if the cubic box). Your Way: deltaP= 2mc, then for the average c= (c^2)/3 ( Which is different than my Prof's) & for Area you just used the Area of one plane not the whole box! But you both got the same result so it's a bit confusing, would you just explain about this part more? Thank you

[@MichelvanBiezen]

Niloufar, It sounds like your professor used all of the molecules in the box and therefore had to consider all three directions. In my explanation, I only took 1/3 of the molecules, so I used one direction only. Since there are so many molecules, it is statistically valid.

[@pipertripp]

Where is the playlist for this series? I can't find it in the vast sea of playlists on your channel. Thx!

[@MichelvanBiezen]

You can find it here: PHYSICS 32 KINETIC THEORY OF A GAS

[@pipertripp]

Thx!

[@lakshminarasimhan7573]

The time of collision is only during impact why take the two times length by Vx

[@MichelvanBiezen]

Because we calculate the pressure on 1 side of the cube.

[@lakshminarasimhan7573]

Michel van Biezen but force is exerted only during the time of collision

[@muaz6424]

@@lakshminarasimhan7573 You could do it that way. You would have a very large force acting over a small amount of time, and then zero force for the entire duration that the molecule would travel back and forth in the box, OR you could average that large force over the entire duration of travel back and forth and you would get the same result: Ft = fT. (his explanation)

[@dishangdoshi9966]

Hello sir would you please explain why does the KE energy of the gas insided a container having piston decreases if we move the piston means while expanding. The question is if the gas is expanding why does the KE decreases and if it does where does the KE goes. Trying to solve it by collisions of gaseous molecules on the wall of container but I am not getting results.. Beacuse if the wall or piston is moving then we cannot apply PV=nRT till the piston stops

[@MichelvanBiezen]

If the gas pushes against the pistorn and moves it, then the gas does work. In order to do work you need energy. That energy can come from the internal energy of the gas, which then cools down as it loses energy. This is accomplished by having the molecules move more slowly.

[@dishangdoshi9966]

@@MichelvanBiezen so if the energy is decreasing then it results that the molecules will move slowly ?

[@dishangdoshi9966]

@@MichelvanBiezen so if the energy is decreasing then it results that the molecules will move slowly ?

[@MichelvanBiezen]

That is correct. The internal energy of a gas is stored by the speed at which the molecules move and at higher temperatures, it can also be stored with molecular rotational modes.

[@deborahgebremariam9358]

thank you so much sir 🙏🙏

[@MichelvanBiezen]

Most welcome 🙂

[@tonmoydeka7319]

Can you upload using polar coordinates

[@asitpurohit6786]

How can you say velocity is equal in all direction ?? What's the reason

[@MichelvanBiezen]

First, we are talking about the root mean square of the velocity. Obviously the velocity of individual molecules is going to vary according to the Botzman distribution. But give that they are at the same TEMPERATURE, their velocities must be the same since their speed depends on the temperature only.

[@lavdieel2369]

But delta t is the time of interaction, not the time of movement! So it is wrong to say that delta t = 2L/vx. Delta t is the small amount of time of the collision between the molecule and the wall. Can you explain me please? I am not getting the idea!

[@MichelvanBiezen]

Yes, that is a concept that many students have questions about. In the macro world, you are correct. But at the quantum mechanic level, the pressure is caused by the number of interactions per unit time and the change of momentum of each interaction. So the calculation in the video is valid.

[@lavdieel2369]

@@MichelvanBiezen Thank you!

[@recall660]

I have a question what is the best book for physics ,for college and university level ,you know some books are too complicated and they don't give you the idea in an understandable fashion ,can you please mention some ,thanks ..

[@conocimientoexpandido]

Love you thanks 🙏

[@MichelvanBiezen]

Thank you and you’re welcome 😊

[@uuitgaurav]

Sir why we used delta t as time between two collison instead of time of contact?

[@MichelvanBiezen]

Because we want to find the number of times each molecule experiences a change in momentum on one side of the container per unit time. Therefore the amount of time the collision takes is not needed.

[@przemysawpierzynowski3284]

I have got question. Why vx=vy=vz ? Is it assumption or we can derive this from some equation? Thank you in advance for reply.

[@MichelvanBiezen]

On average, with the large number of molecules, we can assume that the velocity in all directions will be the same.

[@przemysawpierzynowski3284]

So it is assumption. Thank you for reply. But still, I am going to ask :) because maybe I wasn't precise first time.Is there any chance to derive this from some equation using for example simply newtonian mechanics and limits ?of a function? - when number of molecules and number of collision -> infinity?

[@chufuonlinestudy3074]

sir please can i help what are the essential features of kinetic theory of gases

[@yashvaid7417]

Respected sir, You wrote molar mass is equal to mass multiply with avogadro's number. Can you explain me how ?

[@garekbushnell3454]

Here, when he is referring to mass, he means the mass of a single molecule (or atom, if the gas is made up of unbounded atoms). So if you take a particular molecule's mass, and multiply it by Avogadro's number, you will get the mass of 6.022 × 10^23 of those molecules, which is the definition of molar mass. A concrete example: the molar mass of O2 gas is about 32 g/mol. If you were to divide that mass by Avogadro's number, you should get the mass of a single O2 molecule in grams.

[@raikiri6251]

Great videos as usual! Thanks a lot! ;-)

[@abdullahahmethan2890]

hi, professeur. i don't understand why "change in P=2mv" 2:44 how come the v stays the same before and after collision?

[@MichelvanBiezen]

Abdullah, When the molecule is moving to the left its momentum is: P = - mv When the molecule is moving to the right after the collision its momentum is: P = mv The change in momentum is: Pf - Po = mv - (-mv) = 2 mv

[@talkingfractal]

The collision is said to be 100% elastic, the speed is the same, the velocity just changes direction

[@pohzhenghaoable]

I don’t get why the change is twice explained please?

[@linkhantko150]

molar Mass is gram per mol, I think

[@MichelvanBiezen]

In physics we tend to use metric units

[@Kakashi_Hatake36]

Why isn't the collision time taken? Why do we take the time that needs to travel the length of the box

[@MichelvanBiezen]

We are looking for the contribution of force and pressure for each individual molecule. We know that it collides once every trip across the box and back. We could only consider the collision time and then consider the applied force against the wall zero while it is traveling back and forth, but that would give you the same result and it is easier to calculate it the way shown.

[@Kakashi_Hatake36]

@@MichelvanBiezen sir. I didn't understand. Would please explain in more details

[@MichelvanBiezen]

What part did you not understand?

[@Kakashi_Hatake36]

Why don't you divide the change in momentum by the very little time it takes to bounce off the wall while calculating force. But use the time it takes to travel the length. Why is that time used here

[@MichelvanBiezen]

You could do it that way. You would have a very large force acting over a small amount of time, and then zero force for the entire duration that the molecule would travel back and forth in the box, OR you could average that large force over the entire duration of travel back and forth and you would get the same result: Ft = fT

[@VictoriousSecret1000]

Sir, I like your bowtie. :)

[@recall660]

I will take the chance and ask another question ,is those videos cover all the physics up to university level. For engineering. I mean is this all the physics topics ,thanks prof

[@recall660]

I will take the chance and ask another question ,is those videos cover all the physics up to university level. For engineering. I mean is this all the physics topics ,thanks prof