In a gas, there are lots of molecules traveling at lots of different speeds. Here's a framework for thinking about that.
Log in Nitin 9 years agoPosted 9 years ago. Direct link to Nitin's post “Why isn't V(rms) equal to...” Why isn't V(rms) equal to V(avg) ? Here V(avg) is avg speed of the molecules. • (19 votes) Sean Meadows 9 years agoPosted 9 years ago. Direct link to Sean Meadows's post “Hey Nitin, I think I can ...” Hey Nitin, I think I can answer your question. Let's say there are 6 molecules in a sample of gas. These 6 molecules have the following velocities, respectively: -1 m/s, -2 m/s, -3 m/s, +1 m/s, +2 m/s, and +3 m/s. If we want to find the average speed of a molecule in this sample of gas, we need to convert these 6 vector quantities (i.e. velocities) into 6 scalar quantities (i.e. speeds) by removing the signs. After performing this conversion, we have the following speeds: 1 m/s, 2 m/s, 3 m/s, 1 m/s, 2 m/s, and 3 m/s. To find the average speed, v(avg), we need to add these 6 speeds together and then divide the sum by the total number of speeds: 1 + 2 + 3 + 1 + 2 + 3 = 12 So the average speed is 2 m/s. On the other hand, if we want to find the root-mean-square speed of a molecule in this sample of gas, we don't need to convert those 6 velocities into 6 speeds. Instead, we need to find the square root of the average of the squares of the velocities: (-1)^2 + (-2)^2 + (-3)^2 + 1^2 + 2^2 + 3^2 = 28 So the root-mean-square speed is 2.16 m/s, which is greater than the average speed (2 m/s). This is why the root-mean-square speed does not equal the average speed. I hope this helps. If my reasoning is incorrect, please let me know. (73 votes) Toni Krmek 9 years agoPosted 9 years ago. Direct link to Toni Krmek's post “Hello! Why is it that th...” Hello! • (4 votes) Thom.Wescott 8 years agoPosted 8 years ago. Direct link to Thom.Wescott's post “This is a very good quest...” This is a very good question that gets to the root (pun not intended) of something that is apparently so subtle that it was never explicitly stated to me all the way through second year college physics. Massive bodies in motion have two parameters that diverge the greater their velocity becomes. Momentum is simply mass times velocity, while kinetic energy is one half of the product of the mass and the square of the velocity. So when we look at the simple mode and mean of the Maxwell-Boltzmann distribution (which assumes a uniform mass) we are looking at momentum, but when we consider the rms value we are now looking at a description of kinetic energy. (22 votes) ricky chen 8 years agoPosted 8 years ago. Direct link to ricky chen's post “Hi, I don't quite underst...” Hi, I don't quite understand the idea of "number of molecules per unit speed" on the y-axis. What is this idea of "unit speed"? Which "unit speed" are we dividing the number of molecules by? Could you explain? Thanks! • (2 votes) C Hart 8 years agoPosted 8 years ago. Direct link to C Hart's post “The wording is a little c...” The wording is a little confusing. In order to get a distribution, you are dividing up the molecules into bins according to their speed (like a histogram). In this case, the bins are so thin that instead of a lot of bars, you get a smooth curve that represents the top of each thin bar. So when they say "number of molecules per unit speed" they aren't mathematically dividing, it's more like "number of molecules in each bin of speed" with the bins on the x-axis. But remember that the bins are so thin that we actually refer to them just by the x-value (speed). So there is a number of molecules at 600.000 m/s, and a number at 600.001 m/s, and a number at 600.002 m/s, and thus you get a continuous curve of number of molecules vs speed. (9 votes) Jake Savell 9 years agoPosted 9 years ago. Direct link to Jake Savell's post “Just above the section "W...” Just above the section "What does the area under a Maxwell-Boltzmann distribution represent?", should the 605 m/s quantity be v(rms) instead of v(p)? • (5 votes) Prakriti 6 years agoPosted 6 years ago. Direct link to Prakriti's post “Area under Maxwell distri...” Area under Maxwell distribution curve represents the number of molecules in the system (2 votes) Abhijeeta 8 years agoPosted 8 years ago. Direct link to Abhijeeta's post “Can V(rms) be equal to V(...” Can V(rms) be equal to V(avg)? • (1 vote) ms15135iiserm 7 years agoPosted 7 years ago. Direct link to ms15135iiserm's post “yes when all the velocity...” yes when all the velocity data values are same. (4 votes) Katie 5 years agoPosted 5 years ago. Direct link to Katie's post “If we're plotting speed o...” If we're plotting speed on the x axis, is there then a maximum (the speed of light) the individual particles can't go beyond? So the graph doesn't tail off to infinity? • (3 votes) NoahTheCuber 4 years agoPosted 4 years ago. Direct link to NoahTheCuber's post “Technically yes, but it i...” Technically yes, but it is a bit more complex. As particles approach the speed of light they are called relativistic particles. At that point, special relativity must be taken into account. Look up the Maxwell–Jüttner distribution. It describes the distribution of speeds for relativistic particles. (3 votes) Tatpong 7 years agoPosted 7 years ago. Direct link to Tatpong's post “How did Boltzmann measure...” How did Boltzmann measure gas particle speed? • (4 votes) Shourya M. K. 4 years agoPosted 4 years ago. Direct link to Shourya M. K.'s post “It doesn't. It is just a ...” It doesn't. It is just a representation of particle speed and the number of particles with that speed. In other words, average particle speed is directly proportional to temperature, so you could measure the temperature, but that would be unpractical. (1 vote) overtakou 7 years agoPosted 7 years ago. Direct link to overtakou's post “So what is the interpreta...” So what is the interpretation behind each speed? I read you well explained how the most probable speed is different than the average speed, which are different than the rms speed, and I understood it. But what do they mean physical, for each particle or the gas? • (3 votes) Sukrit J. Baruah 8 years agoPosted 8 years ago. Direct link to Sukrit J. Baruah's post “So, why are there three k...” So, why are there three kinds of mean speeds/velocities? i mean why not just one kind but three. Any reason? • (1 vote) beckydurrant97 8 years agoPosted 8 years ago. Direct link to beckydurrant97's post “It's kind of like how you...” It's kind of like how you'd have a mode, mean and median in statistics. They're all averages, but represent different things. For example, here, the most probable speed is equivalent to the mode, and the average speed is equivalent to the mean. Root mean square is a type of generalised mean, I believe. (4 votes) aebelshajan 4 years agoPosted 4 years ago. Direct link to aebelshajan's post “Why cant we take the abso...” Why cant we take the absolute value for velocity and then use this in our calculations? • (2 votes)Want to join the conversation?
12 / 6 = 2
28 / 6 = 4.67
The square root of 4.67 is 2.16.
Why is it that the average kinetic energy is proportional to the rms speed instead of the average speed .I understood about squares and roots, but still if i have the average speed of particles it would be normal for me to take the average speed for the average kin,energy?
Thank you.
I'm pretty sure you've already understood that, but are still as unsure as I am about why that matters. First off, in the context of this lesson, all we need is to be aware that there is a difference, and that at some point in the future it will become significant.
Looking beyond the scope of this lesson, I expect the implications of this follow from the two facts that a) both momentum and kinetic energy must be conserved, in any non-relativistic event at least, and b) they begin to scale at radically different rates as the velocity increases.
At the human scale, if my car hits another car at 1 mph the momentum is conserved as we both continue at 1/2 mph until various frictions bring us to a stop, and pretty much all of the excess kinetic energy is distributed without too much trauma. However, if I hit the same stationary car at 100 mph and momentum is conserved as we both continue down the road at 50 mph, there is now 10,000 times as much energy to dissipate. My car retains one half times m times 50 squared in its residual velocity, the second car absorbs the same amount as it accelerates, leaving one half m times 10,000 - 2500 - 2500, that is fully half the original kinetic energy to be dissipated as a bunch of highway ugliness.
Just how that translates into behaviours at the molecular scale, I am not sure. Hopefully, I'll get a better handle on this as these lessons progress.
eg. -3,-3,-3,3,3,3 then V aver.: 3 and V rms : 3 .
If any of this is incorrect please let me know and I will correct it.
Thanks!!
