JEE Main Physics Thermodynamics #7 Root Mean Square Velocity

welcome to our lecture online here's a relatively simple problem from the je main test it deals with the root mean square velocities of molecules in a gas and notice the question goes as follows the root mean square speed of molecules of a given mass of gas at 20 degrees celsius and one atmosphere pressure is 200 meters per second the root mean square speed of molecules of the same gas at 120 degrees celsius and 2 atmospheric pressure is x over the square root of 3 meters per second the value of x will be and of course they put the answer in this format since we're not allowed to use calculators on the test makes it a little bit easier to come up with the answer they do throw some information in there that maybe messes you up and the information is that we go from one atmosphere pressure to two and it turns out that makes no difference at all that doesn't come into the equation at least not directly the reason is that the rms velocity vrms is equal to the square root of 3 k t over m if you use m as the mass of a single molecule and k is the gas constant divided by avogadro's number or it can also be written as the square root of 3 r t over the molar mass but again it's essentially the same equation r of course being the gas constant so what we can see here since 3 k and m or 3 r and m since those are constant in the gas and notice there's no mention of the pressure at all so a change in pressure doesn't directly affect the rms velocity we can say that vrms is proportional to the square root of the temperature which means we can then set up ratios we can then say that v1 the initial velocity over v2 is equal to the square root of t1 divided by the square root of t2 since we have that proportionality we can set up the ratios like that now plugging in values remember that t1 was equal to 27 degrees celsius which is equal to 300 kelvin we have to add this to 273 to get 300 and t2 which is 127 degrees celsius will therefore be 400 kelvin so we can plug those numbers in for t1 and t2 and now for v1 we have 200 meters per second so we end up with 200 divided by v2 is equal to the square root of 300 divided by the square root of 400. all right now we can simplify that just a little bit by saying that this is equal to the square root of 3 times the square root of 100 divided by the square root of 4 times the square root of 100 of course this cancels out and we're left with the square root of 3 over the square root of 4 or the square root of 3 over 2. so then the equation essentially becomes 200 divided by v2 is equal to the square root of 3 over 2 and then we can see that v2 therefore is equal to 400 divided by the square root of 3. now we look at the answer format and we see that it's x over the square root of 3 meters per second so therefore we can see that x must equal 400 and that is then the correct answer for this problem so if you remember that equation and again it's all about remembering the equations if you don't remember the you're simply stuck if you remember the equation it's relatively straightforward and i bel i begin to see that more and more the lesson learned here is that you just simply have to memorize all of the equations associated with all the topics in physics if you're going to do well on this test if you don't do that you don't have much of a chance to do well on this so memorization of all of the equations and then on top of that memorization of some of the constants the units of the various components of the equations all that becomes important as you take in the main the je main test as well as the jee advanced test so spending time on that to have all that memorized is a big deal for this test and that is how it's done you