Liquid helium boils at 4.2 K. In a flask, the helium gas above the boiling liquid is at the same temperature. What are (a) the mean free path in the gas, (b) the rms speed of the atoms, and (c) the average energy per atom?

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Hi, everyone. In this breakfast problem, we're being asked to calculate the root mean square speed of the atoms, the mean energy per atom and the mean free path and the gas, we will have a container at atmospheric pressure where the temperature of the krypton gas above the liquid krypton is at krypton's boiling point of 100 and 20 k fin. The atomic mass of krypton is 83.798 U and the atomic radius of krypton is 8.8 times 10 to the power of negative 11 m. We're being asked to calculate first, the root means square speed of the atoms. Second theme energy per atom and third, the mean free path in the guess the options given listing these three different values for our system. And let's jump right into answering our question. First, let's start with the root mean square speed. So the root means square speed of the atoms will be given by the equation of V R MS equals to the square root of three K B multiplied by T divided by M K B is the Boltzmann constant D is the temperature and M is the atomic mass of krypton. It is given in the problem statement that the M or the atomic mass of krypton is 83.798 U. In this case, we have to convert the U into kilograms so that we can calculate everything in si. So doing that, we have to multiply our 83.798 with 1.661 times 10 to the power of negative 27 kg. So this is uh the conversion from U to kilogram. So that our uh total M or our final m atomic mass of krypton and kilogram will be 1.392 times 10 to the power of negative 25 kg. So we can actually next in uh substitute this value into our V R MS to actually calculate our V R MS. Because other than the M, the T is already given at 100 and 20 Kelvin and the K B or the Boltzmann constant will be equals to uh 1.38 times 10 to the power of negative 23 Jules divided by Kelvin. So let's substitute everything into our V R MS formula where V R MS will equals to the square root of three K BT divided by M which will equals to the square root of three multiplied by 1.38 times 10 to the power of negative 23 Joles divided by Kelvin multiply that with the temperature which is 100 and 20 Kelvin and then divide everything with our M or atomic mass of krypton, which will be equals to 1.392 times 10 to the power of negative 25 kg. Calculating this, we will get our V R MS value to then be equals to 100 and 88.9 m per second or rounding it up 100 and 89 m per second. That's the first part of this problem statement. So now we want to move on to the second part when where it is asking for demean energy per atom. So demean energy per atom is going to be given by the equation of E A V or E average A V G. Let's just put it that way equals to three divided by two multiplied by K B multiplied by T. So we can immediately calculate this because we have all the uh information need needed. So E A A V G or E average will then be equals to three divided by two multiplied by 1.38 times 10 to the power of negative 23 Joles divided by Kelvin, which is the Boltzmann constant multiplied that with the temperature which is 100 and Kelvin given in the problem statement, this will give us our E A V G value to then be equals to 2.48 times 10 to the power of negative 21 Jules which will then be the answer to the mean energy per atom. So lastly let's move on to the mean free path in the gas. So the mean free path in the gas is going to be calculated using the following formula where lambda, which is going to be the mean free path in the gas to be equals to one divided by four, multiplied by a square root of two, multiplied by pi multiplied by N divided by V and then multiplied by R squared R is the atomic radius of krypton and N divided by V is the gas number density. The atomic radius of krypton or R is given in the problem statement to be equals to 8.8 times 10 to the power of negative m. So I'm just gonna write that down here. So it's easier for us to see 8.8 times 10 to the power of negative 11 m. The gas number density or N divided by P can be found using the ideal gas law. So let's employ that. So the ideal gas law, according to the ideal gas law, uh N divided by V will be equal to P divided by K B multiplied by T. So this is the substitution that we are going to employ and N divided by V can then be equals. So let's just calculate this automatically. The P is just the atmospheric pressure of 1.13 times 10 to the power of five pascals. So we want to substitute that into our N divided by V value. Our K B is still the bolts one constant and T is still the temperature. So then our N divided by V will equals to 1.13 times 10 to the power of five pastels divided by K B which is going to be 1.38 times 10 to the power of negative 23 Joles divided by Kelvin. And then our temperature will still be 100 and 20 Kelvin. This will be uh give the N divided by V value for our problem to then be equals to 6.1171 times 10 to the power of 25 m cube. Awesome. So now we actually have all the necessary value to calculate our lambda or the mean free pad in the gas. So let's actually calculate that. So our lambda will then be equals to one divided by four multiplied by squared of two multiplied by pi multiplied by and divided by V multiplied by R squared. And then substituting the values we will then get our lambda to then be equals to one divided by four divide multiplied by squared of two multiplied by pi multiplied by the N divided by V which is 6.1171 times 10 to the power of m cube. And then the er squared value which is 8. times 10 to the power of negative 11 m squared just like. So and lastly, what we need to do next is literally just calculate. So calculating this, we will get our lambda value to then be equal to 1.19 times 10 to the power of negative seven m. So that's the final answer for the mean free path of the gas, which is 1.19 times 10 to the power of negative seven m. And then the um mean energy per atom is calculated to be 2.48 times 10 to the power of negative 21 Joles. And then lastly, the root mean square speed of the atoms is calculated to be 100 and m per second. So all of this answer will correspond to option C in our answer choices. So option C 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 our website and that will be it for this one. Thank you.

Liquid helium boils at 4.2 K. In a flask, the helium gas above the boiling liquid is at the same temperature. What are (a) the mean free path in the gas, (b) the rms speed of the atoms, and (c) the average energy per atom?

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Key Concepts

Mean Free Path

Root Mean Square (RMS) Speed

Average Energy per Atom