The closed cylinder of FIGURE CP18.74 has a tight-fitting but ...

Question

The closed cylinder of FIGURE CP18.74 has a tight-fitting but frictionless piston of mass M. The piston is in equilibrium when the left chamber has pressure p₀ and length L₀ while the spring on the right is compressed by ΔL. Suppose the piston is moved a small distance x to the right. Find an expression for the net force (Fₓ)_neton the piston. Assume all motions are slow enough for the gas to remain at the same temperature as its surroundings.

Accepted Answer

Hello everyone. Let's go through this practice problem. A cylinder with a tightly held and frictionless piston of mass. 1 kg is divided into two chambers by a fixed partition. The left chamber has a length of 20 centimeters and contains an ideal gas at a pressure of one atmosphere and a temperature of 25 degrees Celsius. The right chamber contains a spring with a spring constant of 500 nuance per meter initially compressed by five centimeters. If the piston is moved 1.5 centimeters to the right, find the net force on the piston assuming the gas temperature remains constant. Option, a 117 new ones, option B 32.5 new ones, option C 84.8 newtons and option D 52.3 newtons. This is going to be a multi stepp problem. But fortunately, the premise is pretty simple. We want to find the net force acting on this gray piston. So there are two different forces that we're going to want to consider. There is the force acting to the right on the piston due to the air pressure on the left side of the chamber. But the left chamber of the device and then there's the force acting to the left on the piston from the force of the springs compression. Since these two forces are pointing in opposite directions, as long as we can find the magnitude of both forces, it'll be fairly easy to figure out what the net force is. Let's start by determining the force that acts to the right. In other words, the force from the pressure in the leftmost chamber to figure this out, we'll need to know what the final pressure is on that side of the chamber. After all, the pro the problem talks about the piston moving to the right, which means that the gas must be expanding somehow to compare the properties of that gas. Before and after the expansion, we'll use the combined ideal gas law which states that an initial pressure multiplied by an initial volume divided by an initial temperature is constant. So P sub one multiplied by V one and divided by T one is going to be equal to P sub two, a second pressure multiplied by the volume that corresponds to it divided by that temperature. Now, the problem explicitly tells us that the gas temperature remains constant, which means that T sub one is equal to T sub two. So if we ignore the TS in this equation, we're left with a much simpler relationship telling us that P one V one is equal to P two V two, two we're trying to solve for the final pressure. Then let's algebraically solve this equation for P sub two by dividing both sides of the equation by V sub two. So we find that the final pressure is equal to the initial pressure multiplied by the initial volume divided by the final volume. Before we can solve for the final pressure, we'll first need expressions for the two different volumes. We'll need to have those first. So the initial volume, because of the fact that the volume of a cube or rectangular prism is equal to the surface area of the cross section multiplied by the length will say that the initial volume is equal to the initial length of the chamber. L knot multiplied by the chamber's cross sectional area. A Now the problem gives us both of these values in the problem. We're told that the length of the chamber and we can see it in the diagram is centimeters initially. So that's 0.20 m multiplied by a cross sectional area of nine cubic centimeters. Once again, to keep our units consistent, we'll convert this into cubic meters. So instead of nine cubic centimeters, that'll be nine multiplied by 10 to the power of negative four cubic meters. If we put that into a calculator, we find a value for the initial volume of about 1. multiplied by 10 to the power of negative four cubic meters. Now let's find a numerical representation for V sub two, the final volume in the chamber after its length has been expanded by 1.5 centimeters due to the piston moving 1. centimeters to the right. So we'll use the same basic idea, except the length of the chamber at this time is going to be L knot plus X. And that whole term in parentheses is being multiplied by the cross sectional area. So once again, let's compute this. So that's L not, that's 20 centimeters plus the expansion term that 1. centimeters. So that's 21.5 centimeters or 0.215 m multiplied by the exact same cross sectional area. Nine multiplied by 10 to the power of negative four cubic meters. And if you put that into a calculator, we find a final volume of about 1. multiplied by 10 to the power of negative four cubic meters. Now that we have both of our volumes, let's put this into the pressure two formula we found earlier to solve for the final pressure. So first off the initial pressure P sub one is given to us in the problem as being one atmosphere. So it's atmospheric pressure or in past Kells 1.013 multiplied by 10 to the power of five past scales. And this is being multiplied by V sub one which we have established already is 1.8 multiplied by 10 to the power of a negative four cubic meters. And all of this is being divided by the final volume of these sub two, which we found as well to be 1. multiplied by 10 to the power of a negative four cubic meters. We put this into a calculator. We find a final pressure in the right chamber of about 94. 26 multiplied by 10 to the power of three pascals. And now that we found that pressure, we can determine the force coming in from the right. I'll call it force. I'll call it F sub R by taking this pressure and multiplying it by this area of the piston is that is how you determine a, a force from a pressure. So P sub two multiplied by the area. A, this is the pressure. We found 94. multiplied by 10 to the power of three past Kells multiplied by the cross sectional area of the piston. Nine multiplied by 10 to the power of negative four cubic meters, put this into a calculator and we find a force of 84 point 8093 new ones. All right. So that's the force coming in from the toward the right side from the gas. Now, we need to find the force moving toward the left direction from the force of the spring. And recall that when we have a spring force recall hooks law which dates which states that the magnitude of the force is proportional to the spring constant and the compression distance of the spring or in other words, the spring constant K multiplied by the compression distance X. Now we're given the spring constant for the spring. In this problem, it's 500 newtons per meter. And as for the compression distance, well, the problem explicitly tells us that in the initial state of the spring, it is already being compressed by five centimeters. And then after the gas expands, it is then going to be compressed further by an additional 1.5 centimeters. So the total compression length that is going to contribute to this Huella force is going to be equal to the five centimeters or 0. m plus the additional 1.5 centimeters or 0.015 m. If we put that into our calculator, we find a magnitude of the force toward the left of about 32.5 newtons. So now we have the magnitudes of the forces acting in either direction. So if you want to find the net force, all we have to do is take the larger force and then subtract from it, the smaller force. So 84.8093 nuances minus 32 oops, 32.5 nuances is equal to approximately 52.3 newtons. And so that is our answer to this problem. And if we look at our multiple choice options. We can see that this agrees with option D which says 52.3 newtons. So that is our answer to this problem and that is it for the problem. I hope this video helped you out if it did and you need more help. Please consider checking 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.

Key Concepts

Equilibrium in Fluid Mechanics

Ideal Gas Law

Net Force and Newton's Second Law

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