Root Mean Square Velocity Made Easy | Formula, Derivation & Calculations

Hinzugefügt: 2026-05-12

[00:00:00]In this video, I will discuss root mean square velocity.

[00:00:07]If this is a sample of a gas, in it, there are many molecules moving in random directions with different velocities.

[00:00:22]So, it means the velocity of one molecule is different from the velocity of another molecule.

[00:00:29]Now, to describe how fast these molecules are moving as a whole, we cannot use the velocity of a single molecule because it cannot represent the entire gas.

[00:00:42]Well, we may think of finding the average or mean velocity, but unfortunately, that will not work.

[00:00:52]This is because the molecules move randomly in all directions. So, their average velocities cancel out, giving an average velocity of zero.

[00:01:07]So, this does not tell us how fast the molecules are actually moving.

[00:01:12]Now, to describe how fast the molecules are moving, we need a quantity that does not cancel out like the average or mean velocity.

[00:01:25]Now, instead of using the velocity directly, we can actually consider the square of the velocities of all the molecules, then find the mean, then find the square root of that.

[00:01:47]This leads to a quantity called the root mean square velocity.

[00:01:55]So, V R M S is the square root of the mean of the square of the velocity.

[00:02:07]That is why it is called root mean square velocity.

[00:02:14]So, this is the formula. So, what we are going to need now is this V squared mean.

[00:02:23]Ultimately for us, we can obtain this V squared mean from the kinetic energy equation.

[00:02:37]We know that the kinetic energy of a molecule is 1/2 m V squared.

[00:02:47]The average kinetic energy of all the molecules is 1/2 m V squared mean.

[00:03:02]This V squared mean is the mean of the square of the velocity, which is actually what we need.

[00:03:11]Now, from the kinetic theory of gas, the average kinetic energy per molecule is also given by 3/2 k B T.

[00:03:31]So, we can equate this to this.

[00:03:36]That will give us 1/2 m V squared mean equals to 3/2 k B T.

[00:03:49]We can then make V squared mean the subject of the formula.

[00:03:56]Let me clean this.

[00:04:09]V squared mean be equal to 2 * 3 * K B T over M * 2.

[00:04:31]2 will cancel 2.

[00:04:33]So, we have V squared mean equals to 3 K B T over M.

[00:04:45]So, this is what we need.

[00:04:50]We can now go back to the equation of the root mean square velocity.

[00:05:03]Which is V RMS equals to the square root of the mean of the square of velocity.

[00:05:14]So, replace this with this.

[00:05:18]That will give us 3 K B T over M.

[00:05:27]So, that is the formula for the root mean square velocity.

[00:05:34]K B is the Boltzmann constant.

[00:05:38]T is the absolute temperature in Kelvin. While M is the mass of one molecule in kilogram.

[00:05:59]Take note of that.

[00:06:03]There's another fashion of this formula.

[00:06:09]And it is Vrms equals to root 3 RT over capital M.

[00:06:20]R is the molar gas constant. T is still the absolute temperature in Kelvin.

[00:06:24]While this M is the molar mass in kilogram per mole.

[00:06:38]in kilogram per mole.

[00:06:42]Take note of that. It's very important.

[00:06:44]The M in this one is the mass of one molecule.

[00:06:49]While the M capital M here is the molar mass in kilogram per mole.

[00:06:55]kilogram per mole. Take note of that.

[00:06:58]So, that is the formulas for the root mean square velocity.

[00:07:06]This is a question. What is the root mean square velocity of chlorine gas at 25° C?

[00:07:16]Root mean square velocity is root 3 RT over M.

[00:07:29]R is 8.314 J/mol.

[00:07:36]Kelvin.

[00:07:40]T is the temperature and it must be in Kelvin. So, 25°C is equal to 25 + 273 K. And that is 298 K.

[00:07:53]For the molar mass, the atomic mass of chlorine is 35.5.

[00:08:01]So, the molar mass of chlorine gas, which is Cl2, is 35.5 * 2, and that is 71 g/mol.

[00:08:15]But, we must convert it to kg/mol.

[00:08:18]To do this, just divide by 1,000.

[00:08:22]So, that will give you 0.071 kg per mol.

[00:08:32]Take note.

[00:08:34]The molar mass here must be in kg/mol.

[00:08:38]This is very, very important.

[00:08:42]So, we can now go on to calculate the root mean square velocity. That is 3 * 8.314 * 298 over 0.071.

[00:08:57]0.071.

[00:09:00]That is 324 m/s.

[00:09:08]That's the root mean square velocity of chlorine gas at 25°C.

[00:09:15]This is another question. A sample of oxygen has a root mean square velocity of 480 m/s at 27°C.

[00:09:24]Calculate its root mean square velocity when the temperature is increased to 127°C.

[00:09:34]Now, we know that root mean square velocity is root 3 RT over M.

[00:09:46]From this root mean square velocity is directly proportional to the root of the absolute temperature. So, V R M S is directly proportional to root T.

[00:10:05]So, V R M S is equal to K root T.

[00:10:13]Which means V R M S over root T is constant.

[00:10:24]So, the ratio of the root mean square velocity to the root of the absolute temperature is constant.

[00:10:33]You can use this to compare the velocity in the two cases.

[00:10:42]So, this will be V1 over root T1 equals to V2 over root T2.

[00:10:55]V1 here is the root mean square velocity one.

[00:10:59]T1 here is the absolute temperature one.

[00:11:03]V2 here is the root mean square velocity two and T2 is the absolute temperature two.

[00:11:11]You can use this to calculate the velocity when the temperature is increased to 127 degree Celsius.

[00:11:23]Let me clean this.

[00:11:33]Mhm.

[00:11:35]>> V1 is 4 80 m/s.

[00:11:41]T1 is 27° C is what we have to convert to Kelvin.

[00:11:48]That's 27 + 273 and that is 300 Kelvin.

[00:11:54]V2 is what we want to find.

[00:11:57]T2 is 127° C and that is 127 + 273 Kelvin. That is 400 Kelvin.

[00:12:11]So, V1 over root T1 equals to V2 over root T2.

[00:12:19]So, 480 over root 300 equals to V2 over root 4 100.

[00:12:35]I'll make V2 the subject of the formula.

[00:12:39]V2 equals to 480 times this one root 400 over root 300.

[00:12:56]Okay, right.

[00:12:57]V2 equals to 480 times root 400 over root 300 is root 400 over 300 and that is 4 over 3.

[00:13:11]So, 480 times root 4 over 3 is 500 and 54 m/s.

[00:13:21]So, V2 is 550 4 m/s.

[00:13:28]That is the root mean square velocity when the temperature is increased to 127 degrees Celsius.

[00:13:40]This is another question.

[00:13:42]The root mean square velocity of gas at 27 degrees Celsius is UX.

[00:13:47]If the temperature of the gas is increased to 627 degrees Celsius, what is the root mean square velocity of the gas?

[00:13:55]This is similar to the previous question, but the new velocity will be in the form of UX.

[00:14:04]Now, we know that root mean square velocity one over root T1 equals to root mean square velocity two over root T2.

[00:14:17]V1 here will be UX.

[00:14:20]T1 is 27 degrees Celsius, and that is 300 Kelvin.

[00:14:26]T2 is 627 degrees Celsius, and that is 900 Kelvin.

[00:14:34]So, UX over root 300 equals to V2 over root 900.

[00:14:46]Let us make V2 the subject of the formula.

[00:14:49]V2 equals to UX times root 900 over root 300.

[00:15:01]And that is UX times root 900 over root 300 is root 900 divided by 300, and that is three. So, the new velocity will be UX times root three.

[00:15:22]Actually, since the absolute temperature is triple, the velocity will increase by a factor of three since the root mean square velocity is directly proportional to the absolute temperature.

[00:15:39]If the absolute temperature is doubled, the velocity will increase by a factor of root two. Take note of that. It won't double.

[00:15:48]It will increase by a factor of root two.

[00:15:53]Here is another question. Two gases, hydrogen and oxygen, are at the same temperature of 27° C.

[00:16:01]Calculate the ratio of their root mean square velocities.

[00:16:06]The formula for root mean square velocity is root three RT over M.

[00:16:18]At the same temperature, the root mean square velocity of each gas is inversely proportional to the square root of the molar mass.

[00:16:28]So, V R M S is inversely proportional to the square root of the molar mass.

[00:16:38]At the same temperature, so the two gases are at the same temperature.

[00:16:45]Now, we can introduce the proportionality constant.

[00:16:48]So, V R M S equals to K over root M.

[00:16:56]So, we can multiply V R M S times root M equals to K.

[00:17:06]So, the product of root mean square velocity and the square root of the molar mass is constant at the same temperature for each gas.

[00:17:21]Which means Vrms1 root M1 equals to Vrms2 root M2.

[00:17:34]One is for hydrogen, two is for oxygen.

[00:17:39]You can rearrange V1 over V2 equals to root M2 over root M1.

[00:17:55]And that is equal to root M2 over M1.

[00:18:04]So since one is for hydrogen and two is for oxygen, it means V hydrogen over V oxygen equals to root M oxygen over M hydrogen.

[00:18:30]Using this formula, you can calculate the ratio of their root mean square velocities.

[00:18:37]So, let us do that.

[00:18:56]The molar mass of oxygen is 32 g per mole, while that of hydrogen is 2 g per mole.

[00:19:10]You don't need to convert to kilogram per mole because it involves division, so the unit will still cancel out even if you convert.

[00:19:16]But in actual calculation of the root mean square velocity, you have to use kilogram per mole, but here you don't need to convert because the unit gram per mole gram per mole will still cancel out. If it is kilogram per mole, it will still cancel out, so it will give you the same thing.

[00:19:33]So, V H2 over V O2 equals to root M O2 that's 32 over M H2 that's two.

[00:19:47]So, V H2 over V O2 equals to root 32 / 2 is 16.

[00:19:55]So, root 16 and root 16 is four.

[00:20:02]So, V H2 over V O2 is equal to four.

[00:20:05]Which means the ratio V H2 ratio V O2 is four ratio one.

[00:20:20]From this, it means V H2 is four times V O2, which means hydrogen gas moves four times faster than oxygen gas at the same temperature.

[00:20:35]That is the meaning.

[00:20:37]Hydrogen gas moves faster four times faster than oxygen gas at the same temperature.

[00:20:48]So, that is the solution to the question.

[00:20:53]That is the end of this video.

[00:20:56]Thanks for watching.