Equation 20.3 is the mean free path of a particle through a gas of identical particles of equal radius. An electron can be thought of as a point particle with zero radius. b. Electrons travel 3.0 km through the Stanford Linear Accelerator. In order for scattering losses to be negligible, the pressure inside the accelerator tube must be reduced to the point where the mean free path is at least 50 km. What is the maximum possible pressure inside the accelerator tube, assuming T = 20℃? Give your answer in both Pa and atm.

Video transcript

Hey everyone. Let's go through this practice problem. A beam of beta particles passes through a cylindrical chamber with a radius of 11.5 centimeters and a length of 80 centimeters filled with nitrogen gas. If the mean free path of the beta particles is measured to be one centimeter, what should be the pressure inside the chamber? When the temperature is at negative one degrees Celsius, let us assume the nitrogen atoms have a radius of 0.1 nanometers option A 6.2 pascals, option B 8.5 pascals, option C 10 pascals and option D 12 pascals. Since we're looking for the pressure, we can relate it to other thermodynamic properties by using the ideal gas law and one of its forms states that the pressure of the gas multiplied by its volume is equal to the number of particles multiplied by the Voltz constant multiplied by the temperature of the gas. We're solving for pressure here. So if we solve for pressure algebraically, by dividing both sides of the equation by V, we can see that pressure is equal to the number of particles divided by the volume of the gas multiplied by the boltzmann constant and multiplied by the temperature. What we're missing, however, is the number of particles N and the volume of the gas V. And we'll need either both of those to plug it into the equation or we'll just need to know what the end to V ratio is. And we can find that end to V ratio by using the formula for the mean free path of the particles, which is actually given to us in the problem recall that the formula for the mean free path is equal to the Boltzmann constant multiplied by the temperature of the gas divided by the square root of two multiplied by pi multiplied by the square of the radius of the particles multiplied by the pressure of the gas. If we're going to use this to solve for N divided by V. Well, first, we want to find a relationship between the end to V ratio and the variables we have in the problem. Let's actually take the ideal gas formula we found earlier and resolve it for N divided by V algebraically, we can see that the end to V ratio is equal to the pressure divided by the Boltzmann constant. And the temperature, we can now work this into our mean free path formula by substituting N divided by V N four P divided by K T. And what we find is that our new formula for the mean free path is equal to the volume divided by the square root of two pi R squared N very simply algebraically solving this equation for the ratio N divided by V, we find that that ratio is equal to one divided by the square root of two pie R squared multiplied by the mean free path of the beta particles. So to find the end of E ratio, we just, we now have all the variables we need the squared of two pi R squared is just the square of the radius of the nitrogen atoms. So that's given to us as 0.1 nanometers or alternatively in meters 0.1 multiplied by 10 to the power of negative nine m. Don't forget to square it. And then it's multiplied by the mean free path of the beta particles which is given to us as one centimeter or 10.1 m. We put this into a calculator. We find that our end of E ratio is equal to about 2. multiplied by 10 to the power of 21 particles per cubic meter. Now let's revisit the pressure formula we discussed well earlier on in the problem and plug in the new value we found into that. So recall that we discovered pre P pressure to be N divided by V multiplied by K T. We now know this to be 2.25 multiplied by 10 to the power of 21 particles per cubic meter. This is multiplied by the Boltzmann constant 1.38, multiplied by 10 to the power of negative 23 jewels per Kelvin and then multiplied by the temperature. And the temperature is given to us in the problem as negative one degrees Celsius, we have to convert this into Kelvins by adding 273 which we're adding 273 to a negative one. So pretty simply that's just 272 Kelvins. And if we plug this into a calculator, then we find a pressure of approximately 8.5 pascals. So 8.5 pascals is our answer to this problem. And if we look at our multiple choice options, we can see that this is exactly what option B says. So option B is the correct answer to this problem and that is all for this video. I hope this video helped you out. If you feel you need more practice, please check out some of our other videos which will give you more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.