Oxygen (O2) has a molar mass of 32.0 g/mol. What is (g) How many ...

Question

Oxygen (O2) has a molar mass of 32.0 g/mol. What is (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm?

Accepted Answer

Hey, everyone today, we're dealing with the problem about pressures and forces. So we're being told that we have nitrogen which has a molar mass of 28.01 g per mole. And we're being told that the root mean square speed of nitrogen is 509 m/s. With this in mind, we're being asked to find how many molecules with the speed moving between the left and right faces of a cube that has a length of two m or 0.2 m are required to produce a pressure of 1.5 atmospheres. So the first important thing to note here is that we're doing between the left and right faces. So what does this mean? We're dealing with a one D pressure problem? We're only working in one dimension, were not working in a full three dimensional cube. We're just considering the left and right faces and particles bouncing off. So uh to go ahead with this problem, we need to do a few things and we'll go step by step. So the first step is to find the root mean squared speed, Which in this case is already given to us. So we didn't have to calculate it or anything. It's just 509 m/s. The second step is to determine the change in momentum when a molecule collides of the wall, assuming that all speed is perpendicular to the area. So let's go ahead and think about that. We're being asked to or we have to find the change in momentum and to do this, we can utilize the impulse theory. So recall that the info theorem states that and impulse and impulse, let me write this a little higher and impulse is equal to force into time, which is equal to just the change in momentum, which can also be written as mass times the change in velocity. So that in mind, let's go ahead and look back and determine the change in momentum, right. So we can say that since we're dealing with, if I, if I was to draw this out for you guys, it would be a sort of these are the two sides accused and we're dealing with this sort of lateral movement of a particle bouncing between the two. So we're dealing with the X direction. So we can say that delta P in the X direction is equal to or sorry mass times the change in the horizontal philosophy. So we have that. So what is changing velocity? Well, that can also simply just be written as mass into the final velocity in the X direction minus the initial velocity in the x direction. Now velocity is a magnitude or sorry, the velocity magnitude is conserved when a molecule collides with the wall, right. So in the case of the wall, we can simply say that the initial velocity will be equal to or the magnitude of the initial velocity will be the same as the magnitude of the final velocity, right? But since we're dealing with a sort of horizontal movement, we can say that if this is the center, any movement to the left will be negative and any movement to the right will be positive. In terms of sine, the magnitude doesn't change, but the sign does, it indicates our direction. So a collision on the left wall would be negative, let's say the collision of left wall will be um ah the initial velocity. So it'll be negative collision of the right wall will be positive simple as that. So if we use that logic, we can go ahead and put that back into our formula that we just wrote. And we can say that off the left wall of the initial velocity because the entire period that we're concerned with that we're going to need the entire interval will be a particle starting on the left side, the initial going to the right and then bouncing back and returning to his initial point that will be the whole cycle. So with that in mind, we can also go ahead and say that this momentum. The change of momentum can also be ah mass times the initial velocity minus the inverse of the initial velocity because the signs will be different, right? Who said that their magnitudes are the same? And if initial velocity is negative, and that means that having sorry, I wrote this incorrectly, this is final my bad. So we have the final velocity minus the initial velocity and the initial velocity is negative. So this can be therefore be written as um which is mass and kilograms is V F minus X plus the initial velocity. So with this, we can also say that the uh since the magnitude doesn't change, we can also therefore say that this will be the absolute value, the magnitude of the root mean squared velocity will also be the same as the other two velocities. I'd say that's a safe assumption to make. So that in mind, we can go ahead and plug that back into our equation. And we can say that using that proof, this is therefore equal to two mm V not X, which can also be written as two M V RMS because V rms is the same thing as the initial velocity in terms of magnitude. So from here, the next thing we need to do is to determine force using the impulse theorem that I mentioned earlier. Again, this is our impulse theorem right sir comms this is our impulse theme right here. And it tells us that a force multiplied by time is equal to change in momentum. So rearranging for force, we can say that force is equal to change the momentum over time. Now, what is time here? Well, it's not mentioned, but again, time is the interval that it takes as I mentioned earlier that it takes for a particle to go from the left wall to bounce off the right wall and then come right back to its initial starting point. The time it takes to complete one full round of travel, let's say two bounces off the right wall back to the left wall. So with that, we can also say that the distance, the distance for time. So if, and I'll write this in blue, we can say that the time is equal to the distance over the velocity, right, that is the definition of time. And we know that the vote and the distance will be two times the length of the cube because we're going one length and then back that same distance. So it'll be two L over V RMS. And we know that length Is listed as 0.2 m. So substituting all of this back into our force equation, we can say that we know that change of momentum is equal to two M the RMS right. And if you're dividing by time, well, that'll just be the inverse, it'll just be multiplied by one over T which we can also rewrite as two M V R M s multiplied by V R M S over two L because it's the inverse of time which we determined to be this simplifying this and factoring, we get that force will be equal to M into V R M S squared. Oops. Nope, that should be outside the parentheses over to L. So that is our apologies not to L just L because our twos will cancel out here. So with this in mind, we now have the formula we need to solve the force. So the next thing to do is to determine pressure using uh force divided by area, right, for just one molecule. And we'll do this on just one of the faces for now. So let's write this out over here. So we know that pressure is equal to force divided by area, which we can say, since we're dealing with the square, we're dealing with the Q, which means one side of the one side will just simply be L squared. And we've determined that force is equal to the term that we discovered earlier. So let's substitute that in, we can say M V R M S squared over L multiplied by one over squared. Which means if we simplify, we then get um it's played by V RMS square over L Cute. So that's our formula for pressure. Now, the last thing we need to do is convert the pressure of one molecule to all molecules. And let's determine how we're going to do that. So real quickly, I'm going to scroll down just a little bit. We have a little more space. Let me write and let me put this a little more in frame. So our last step is to convert the pressure for one molecule to all molecules again. So to do this, to do this, we can go ahead and right that we're looking for the pressure of all molecules, right? And to do that well, or the pressure of how many molecules we have to say the least and this is an equal time, but we can say that the pressure of a ah or the entire pressure should be equal to and which is avocados number. And remember N is equal to 6.22 times 10 to the 23rd molecules which I represent with MOL M O L E, not to be confused with moles, that's molecules specifically. But so this will be avocados number multiplied by the pressure of a single molecule Q. And were given that the pressure That we're trying to find the total pressure is 1.5 atmospheres. However, you want this in pascal. So let's just sort of convert that real quick. We know that PUX 1.5 A T M. And we can recall that the conversion factor is that we have 1.13 times 10 to the fifth Pascal's for every one A T M Giving us a pressure of 1.5-0 times 10 to the 5th pascal's because our atmospheres will cancel out. Now to find or to find how many mall like how many molecules we have with the speed just between the left and right distances, we need to utilize the mass that were given earlier the molar mass. So we can say that, excuse me, leaving this section aside for the time being. And let me write this in red for you guys, we know that molar mass is simply equal to the polarity divided by, I'm sorry, excuse me, the mass or molar mass, I should say molar mass is simply equal to avocados number multiplied by the mass given which means that mass is therefore equal to the molar mass divided by avocados number, right? And we can substitute uh for m in this term that we're looking for as well. And we want our morality to be in kilograms because we're going to R S I units. That's what we're using for all our calculations. So we can say that this will be 28.01 g Permal right over avocados number which I've written above recall that if we want to convert grams to kilograms, we take 28.1 g per mole, convert this two kg where we can recall that we have for every one kg, we have 10 to the third grams. So this will give us an answer of 28.01 times 10 to the negative three kg which will work here as well. Once we convert to grams to kilograms over, I forgot his number which gives us an answer or whoops my bad. So we have this term not just yet, but we have this term that we can substitute in for mass, right? So finally, and I'm gonna scroll down once more with this, we can go ahead and solve for the, let's just keep our values there. We can solve for the pressure of a singular molecule because we can then use that to again determine the pressure of all the molecules that were already given. So this is the pressure of all the molecules. So we can go ahead And let me just write out a few more things that we're going to need. We can always remember that one pascal is equal to one Newton per meter squared and one Newton is equal to one kilogram kilogram meter per second square. And we're going to need these. So excuse me, the pressure of a single molecule, pressure of a single molecule we determined was M V R M S squared over L cute, which we can now substitute in some values. Now, we said mass is equal to molar mass divided by avocados numbers. So we can substitute that in here, we can say this into V R M S square over avocados number multiplied by L cubed. And now we can substitute in our values. We have 28 01 times 10 to the -3 kg. Multiplied by 509 m per second squared whole squared divided by 6.02, 2 times 10 to the 23rd molecules. Oops wrong bracket multiplied by A distance of 0.2 m Raised to the 3rd power Which gives us an answer of 1.507 times to the negative 18th Pascal's. We're still not done though because again, remember we're trying to find the number of molecules that we're going to need, right? And I apologize. I made a mistake earlier. This end here is not avocados number. It is simply the number of molecules not to be confused with avocados number which I'll know Tate with a little a as such. So with that in mind, now, we can use the values that we've had or that we've discovered to go ahead and find the number of molecules needed to create this pressure, right? So and is equal to the entire pressure over the pressure of a single molecule which we can substitute in our values for we have 1. times 10 to the fifth. Pascal's over 1.507 times 10 to the negative 18th Pascal's which gives an answer of 1.1 times 10 to the 23rd molecules. So 1.0, 1 times 10 To the 23rd molecules. And if we look at our answer choices that lines up with answer choice c one point oh one times 10 to 23rd molecules. Now, quick thing to note before I finish this off is that we're dealing with a one dimensional space here. So this is actually only the pressure between two faces of the cube, between the left and right sides of the cube. So the actual total pressure Or the total number of molecules as well will actually be three times this amount or 3.01 or 3.03 times 10 to the 23rd molecules. However, we're only dealing with the left and right faces of the cube. In this question, we're only dealing with one dimension. So the answer will be C I hope this helps and I look forward to seeing you all in the next one.

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Oxygen (O2) has a molar mass of 32.0 g/mol. What is (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm?

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