5.0 x 10²³ nitrogen molecules collide with a 10 cm² wall each second. Assume that the molecules all travel with a speed of 400 m/s and strike the wall head-on. What is the pressure on the wall?

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Hey, everyone. Let's go through this practice problem. An oxygen cylinder for scuba diving is placed horizontally on the ground. Consider that five multiplied by 10 to the power of 24 oxygen molecules moving at 485 m per second in the X direction collide head on with a 50 square centimeter wall of section of cylinder. Each second determine the pressure exerted by the oxygen on the walled section. Option A 1.1 multiplied by 10 to the power of two pascals. Option B 4.5 multiplied by 10 to the power of two pascals. Option C 5.2 multiplied by 10 to the power of four pascals. And option D 1.7 multiplied by 10 to the power of five pascals. So first off, because this problem is asking about the effects of many little molecules colliding against a surface. Let's first recall the formula for the force experience to buy a surface when a molecule or a particle collides into it. Recall that it's connected to the impulse or change momentum of the particle. So along the x direction, for example, it's equal to two multiplied by the mass of the particle multiplied by its speed in the x direction of the sub X divided by the period of time over which this takes place. However, if we're looking at the motion and the collisions of many, many particles, this is also proportional to the number of collisions that occur. Additionally, if the force of all these collisions is manifest is manifesting itself as the pressure experienced on the tank, then this force has the same magnitude as the force experienced due to the pressure. And recall that force due to pressure is equal to the magnitude of that pressure multiplied by the surface area over which the pressure is occurring. For the purposes of this problem. It's this pressure variable that we want to find. So let's algebraically solve this equation for the pressure P by dividing both sides of the equation by the area A. So the pressure is equal to two M V sub X M divided by the surface area multiplied by the change in time. Note that since the problem gives us the number of oxygen molecules that collide each second. And one of the terms in this equation is the number of collisions divided by the period of time, the interval of time. Then we can equate those two variables together as a rate of collisions per second, which is given in the problem as five multiplied by 10 to the power of four. And that's collisions per second. Another variable we need in order to put this into a calculator is the mass of the particles, we know from the problem description that these are oxygen particles that we're talking about. And if you look at in, at a periodic table of elements, you'll see that oxygen has a molar mass of 16, at 16 atomic mass units. But one oxygen molecule is diatomic. So there are two different atoms. So one oxygen molecule actually has 32 atomic mass units. However, in order to use this in the problem in the equation, we need to convert this into kilograms. So to convert it from atomic mass units to kilograms, you have to multiply this by 1.66 multiplied by 10 to the power of a negative 27 kg. Since that's the mass of one atomic mass unit, this gives us a mass of about 5.321 multiplied by 10 to the power of negative that's negative 26 kg. And now we have everything we need to start putting this into a calculator that's two multiplied by the mass which is 5. multiplied by 10 to the power of negative 26 kg multiplied by the speed in the horizontal direction or 485 m per second, multiplied by the collision rate or five multiplied by 10 to the power of four or 2024 collisions per second, all divided by the area which is given to us as 50 cubic centimeters. But we actually want to write this as cubic meters in order for our units to be consistent. So we'll actually write this as 50 multiplied by 10 to the power of negative four cubic meters. And now, if we put this into a calculator, then we find a pressure of approximately 5. multiplied by 10 to the power of four pascals. And so that is the answer to the problem. And if we look at our multiple choice options, option C agrees with this 5.2 multiplied by 10 to the power of four pascals. So option C is the correct answer to this problem. And that's all for this video. I hope this video helped you out. P please uh if you need more practice, please check out some of our other videos if you'd like more experience with these types of problems, but that's all for now. And I hope you all have a lovely day. Bye bye.

5.0 x 10²³ nitrogen molecules collide with a 10 cm² wall each second. Assume that the molecules all travel with a speed of 400 m/s and strike the wall head-on. What is the pressure on the wall?

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Key Concepts

Pressure

Momentum Transfer

Ideal Gas Behavior