Hey if you are still interested in the derivation, you can go to my site, click on the physics section and go to the chapter that is titled "Kinetic Theory and van der Waal's Equation". The second and third lectures should do the trick.
The root mean square velocity comes from the kinetic molecular theory of ideal gases. I actually show you how to derive it in one of my lectures, so if you are still interested, let me know and I will direct you.
Hey, thanks for noticing ! Someone already noted that and I made a new lecture (actually two). In the first lecture, I derive the relationship between the average translational kinetic energy and temperature using the Kinetic Theory and in the second lecture, I use that to derive the equation for the root mean square velocity. You can find these two lectures on my website under the chapter "Kinetic Theory and van der Waal's Equation" in the physics section (lectures 2 and 3).
I have a question. I was told that the average velocity depends on the molar mass too. Can you please explain this? Thank so much for the videos! They are great :-) You have no idea how helpful you have been!!
Hey if you are still interested in the derivation, you can go to my site, click on the physics section and go to the chapter that is titled "Kinetic Theory and van der Waal's Equation". The second and third lectures should do the trick.
Where is the formula come from
HI SIR I AM FROM INDIA , THANKS FOR SUCH A BEAUTIFUL EXPLANATION , BY THE WAY YOU GAINED A SUBSCRIBER .CHEERS .........
You have made clear in four minutes what my p-chem professor could not with 50. Many thanks!
i am seeing this in 2019, and this was amazing.
The root mean square velocity comes from the kinetic molecular theory of ideal gases. I actually show you how to derive it in one of my lectures, so if you are still interested, let me know and I will direct you.
Hey, thanks for noticing ! Someone already noted that and I made a new lecture (actually two). In the first lecture, I derive the relationship between the average translational kinetic energy and temperature using the Kinetic Theory and in the second lecture, I use that to derive the equation for the root mean square velocity. You can find these two lectures on my website under the chapter "Kinetic Theory and van der Waal's Equation" in the physics section (lectures 2 and 3).
check your statement (B) average kinetic energy= mRT/2M ( Its written wrongly on your white board)
it should be 3mRT/2M
Good explanation in short time. 👍
Great lecture! It really helps to see it all together at once.
Reward this guy🙏
How to derive RMS velocity
You, sir have saved me for me physics paper.
fantastic teaching
Fabulous 🔥🔥
You are very nice teacher, Vav is given in book as root2RT by M
Thanks...solved my query
This was really helpful. Thanks!
I have a question. I was told that the average velocity depends on the molar mass too. Can you please explain this? Thank so much for the videos! They are great :-) You have no idea how helpful you have been!!
That is great to hear!
Thank you!!
you're welcome!
helped me a lot! Thank you so much
Hey Dave. I'd say wait until you complete Calc 2. You need to be familiar with integration and limits.
1.38 × 10^-23 m2 kg s-2 K-1
Great to hear it! :)
what does R equal? is it R= kg*m^2/mol*s^2*K or is it R= 8.314Kg*m^2/mol*s^2*K ?
alright. ill do that, eventually.
How do you put them into a math problem?
what kind of calc does one need to derive the formula? I am in calc I, is this enough?
I dont just understand the difference between the average velocity and the root mean squared velocity can you explain this
Thank you :)
Where does it come from?
i am interested in the root mean square derivation please
How to derived RMS velocity??
What is the value of k?
Thanks!
Sir how to derive root mean square velocity
how to derive rms velocity
This video was made when i was in 3rd grade
can you dreive it for us please sir
Can you please show me how to prove the the average kinetic energy equation
amen!
molar mass should be in kg
Doesn't avg ke depend on f(degree of freedom)
This was so helpful. Thanks!