The mean free path of a molecule in a gas is 300 nm. What will the mean free path be if the gas temperature is doubled at (a) constant volume and (b) constant pressure?

Video transcript

Hi everyone in this practice problem, we are being asked to determine the mean free path of a molecule in a gas if the gas temperature is increased by a factor of four, given that the gas is kept first at constant volume and second at constant pressure, it is known that initially the molecule has a mean free path of 450 nanometer. And the options given for the mean free path or the new mean free path is going to be a 1st 450 nanometer at constant volume. And 2nd 1800 nanometer at constant pressure. B 1st 900 nanometer and 2nd 1800 nanometer c 1st 1800 nanometer and 2nd 450 nanometer, the 1st 450 nanometer and 2nd 900 nanometer. So we are given the mean free path or the initial mean free path of the molecule, which I'm going to represent with lambda one to be close to 450 nanometer. So we're going to let the one to be the temperature of the molecule B one to be the pressure of the molecule and P one to be the volume of the molecule at this particular condition. And we want to recall that the formula for lambda or the mean free path is going to be given by one divided by four squared of two pi and N divided by V also are squared. This actually implies this equation actually implies that lambda is going to be directly proportional to uh P which is volume or through rearrangement, we will get volume or lambda will be equals to volume divided by four multiplied by squared of two, multiplied by pi multiplied by N multiplied by R squared. And through this formula, we can actually look more clearly that Lambda is directly proportional to volume. The new temperature or T two is going to be four times T one which is given in the problem statement and the new volume or P two will be equals to P one considering that it is kept at constant volume. So at the first condition, P two will be equal to P one since P two is equal to P one and Lambda is proportional or directly proportional to V. And when we are looking into Lambda one divided by lambda two, then the equation will then just equal to P one divided by V two. Considering that all of this other for uh this other variables are constant, then we can determine that P one divided by P two will be equals to one. And then that will give us Lambda one to be equals to lambda two, which will be equal to 450 nanometer. And that will be for the first part, for the second part of this following statement at constant pressure we will use or employ the ideal gas equation where P one multiplied by V one divided by N multiplied by K B multiplied by T one will be equal to P two divided by V 22, P two multiplied by V two, divided by N multiplied by K B multiplied by T two. And we want to notice that T two is supposed to be, or it will be equals to four T one. So then this equation will then be equal to P two multiplied by V two divided by N multiplied by K B multiplied by four T one. Awesome. So therefore, at constant pressure or at P one equals to P two, then we can determine from this equation right here. Then uh P one divided by V two or V one uh over V two will be equal to one divided by four or essentially P two will be equals to four V one. And from there, we can determine that Lambda one divided by Lambda two will then be equal 21 divided by four. Since Lambda is directly proportional to P. So since Lambda one divided by Lambda two will be equal to one divided by four. Del Lambda two will be equal to four Lambda one or will be equal to four multiplied by 450 nanometer, which will come up to a value of 1800 nanometer. So that will that will be the answer to this particular practice problem where for the uh case of constant volume, then the new mean free path is going to be the same as 450 nanometer and and the case at the case of constant pressure, then the new mean free pa is going to be 1800 nanometer and these answers will be corresponding to option A in our answer choices, which will be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website and that will be it for this one. Thank you.