At 100℃ the rms speed of nitrogen molecules is 576 m/s. Nitrogen at 100℃ and a pressure of 2.0 atm is held in a container with a 10 cm x 10 cm square wall. Estimate the rate of molecular collisions (collisions/s) on this wall.

Question

Accepted Answer

Hey, everyone. Let's go through this practice problem. Argon gas is stored in a tank with a rectangular surface of sides 15 centimeters and 25 centimeters. The tank is pressurized to three atmospheres and kept at a temperature of Kelvins. The root means square velocity of the Argon gas is found to be 430 m per second. Calculate the number of collisions per second that occur on the rectangular surface of the tank. You have four, multiple choice options. Option A 8.1 multiplied by 10 to the power of collisions per second. Option B 8.8 multiplied by 10 to the power of 24 collisions per second. Option C 3.5 multiplied by 10 to the power of 26 collisions per second. And option D 6.1 multiplied by 10 to the power of 26 collisions per second. When we're dealing with molecular collisions. The key formula to remember is that in one dimension, the force that occurs as a result of the collision is equal to two multiplied by the mass of the particle multiplied by the velocity in that dimension. So B sub X, for example, in the X dimension divided by the period of time over which that collision took place. When we're talking about many, many molecules hitting a wall at once where there are a lot of collisions to work with. This force is also pro proportional to the number of those collisions. N. Since this problem is asking for the number of collisions per second, essentially, what it's asking for is this big N divided by delta T term. That's what we're looking for. The amount of collisions that occur per unit time. So let's algebraically rewrite this equation to solve for big N divided by delta T. So dividing both sides of the equation by two M V, we find that the collision rate is equal to the force on the wall divided by two M V sub X. This equation will give us the collision rate. The problem is asking for, but there are a few other things we need to find first in order for it to work because we're not given the force that the rectangular wall has experienced, but we are given the surface area of the wall. And we're told what the pressure of the tank is. Now when we have a lot of collisions due to a gas or really any type of fluid, the force from those collisions is what manifest is what manifests itself as a macroscopic pressure. So instead of just using a force F, we can instead write the force as a force due to a pressure. And recall that the force due to pressure is equal to the magnitude of that pressure multiplied by the surface area over which the pressure acts. So P A divided by two M V sub X is another way to write the formula we're trying to find. And this is more helpful to us because we have the pressure and the dimensions of the rectangular wall. However, there are two things we're still missing the mass and the V sub X if you want, you could have written that as V sub Y or V sub Z. But the point is we're missing the speed in one of the dimensions. We are. However, given the root means square velocity. And so this is where it's helpful to remember what the formula for root means square velocity is. The R M mass velocity is equal to the square root of the mean of the squares of the velocities or all the different particles that are moving in the gas or writing this out more formally, the sub R MS is equal to the square root of A V sub X squared averaged for all the particles in the fluid plus V sub Y squared averaged plus V sub Z squared averaged. Additionally, since we're looking at a gas where everything's moving around in presumably random dimensions, we can assume that the average speed in the X component, for example, is going to be the same in all three dimensions. So we can assume that V sub Y squared averaged is the same as V sub X squared averaged. And we can assume the same thing about V sub C squared averaged. So the R MS speed can be more simply written as the square root of three multiplied by V sub X squared averaged. If we make some basic assumptions about this gas, like it's an ideal gas and that the average velocity is the same as the true velocity. Then we can even more simply write this as the square root of three multiplied by via sub X because the square root canceled out with the square that was there. So if we want to solve for V sub X, we'll simply divide both sides of the equation specifically V sub R M F S by the square root of three. So the problem tells us that the R MS velocity is 430 m per second. So if we divide this by the square root of three in a calculator, then we find a speed in one dimension of about 240 eight 0.261 m per second. The final thing we need to determine before we can plug our values into a calculator is the mass. We need to know the mass of one of the particles. Well, the problem tells us that we're looking at an Argon gas here. So if you look at a periodic table, we find that Aragon has an, has a mass of about 40 atomic mass units. Well, we need to convert this into kilograms in order to put this into the calculator. So let's use the typical conversion for that. It's about 1.661 multiplied by 10 to the power of negative 27 kg. 41 U put this into a calculator. And we find a mass for one Argo molecule about 6.644 multiplied by 10 to the power of negative 26 kg. So all we need to do now to find the collision rate is plug the values we found into a calculator for the formula we determined earlier of pressure multiplied by area divided by two M V. So the pressure is given to us in the problem as three atmospheres, which means we'll have to convert that into past scales by multiplying it by 1.13, multiplied by 10 of the power of five pascals. We then multiply by the area of the wall which is given to us in the problem as 15 centimeters by 25 centimeters. So of course, in meters, this is 250.15 m multiplied by 0.25 m. None of this is all divided by two, multiplied by the mass. We determined earlier of 6.644 multiplied by of the power of negative 26 kg multiplied by the speed we discovered earlier of about 248 0.261 m per second. So now if we put this into a calculator, we find a collision rate of about 3.5 multiplied by 10 to the power of 26 collisions per second. And that is the answer to a problem. If we look at our options, we can see that this answer agrees with option C. So option C is the correct answer to this problem and that's it for the problem. I hope this video helped you out. If 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 good day. Bye bye.

At 100℃ the rms speed of nitrogen molecules is 576 m/s. Nitrogen at 100℃ and a pressure of 2.0 atm is held in a container with a 10 cm x 10 cm square wall. Estimate the rate of molecular collisions (collisions/s) on this wall.

Verified video answer for a similar problem:

Key Concepts

Root Mean Square Speed (rms speed)

Kinetic Theory of Gases

Collision Rate