The 50 kg circular piston shown in FIGURE P18.57 floats on 0.12 mol of compressed air.

a. What is the piston height h if the temperature is 30°C?

Question

The 50 kg circular piston shown in FIGURE P18.57 floats on 0.12 mol of compressed air.

a. What is the piston height h if the temperature is 30°C?

Accepted Answer

Hello, fellow physicists today, we're going to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. An industrial machine as illustrated below uses compressed 0.23 moles of carbon dioxide gas to lift a square shaped movable piston that weighs 38. kg. The side of the piston has a length of 15.0 centimeters. If the temperature of carbon dioxide gas is 300 Kelvin determine the value of L. So our angle is to find the value of L. So we're trying to find the length. OK. So we're given some multiple choice answers. They're all in the same units and meters. So let's read them off to see what our final answer might be. A is 0.15 B is 0.22 C is 0. and D is 0.40. OK. So first off, let's make the following assumptions. Let's assume that carbon dioxide behaves or carbon dioxide gas behaves like an ideal gas that the process is adiabatic. That the system remains in static equilibrium and that the piston moves slowly enough that the process can be considered quasi static. OK. We also need to recall and use the ideal gas law equation which states that the pressure multiplied by the volume is equal to the number of moles multiplied by the universal gas constant multiplied by the temperature. OK. So we need to rearrange this equation. We need to rearrange the ideal gas equation to solve for B. So when we do that using algebra, we need, we will get that the volume is equal to the number of moles multiplied by the universal gas constant, multiplied by the temperature all divided by the pressure. We also need to solve for pressure. So we need to recall and remember the equation for the pressure of a piston. So the pressure of the piston. So the equation to find the pressure of a piston is as P is equal to the pressure of the atmosphere plus the pressure of the piston. Oh But also we need to take into account, we need to find out how to find the pressure of a piston. So let's rewrite this equation and recall how to find the pressure of a piston. So the pressure of a pi of the piston equals P E equals pressure atmosphere plus the wheat divided by the area. OK. So let's make a couple of notes here that the weight is equal to the mass multiplied by gravity. And that the area which in this case, we're dealing with a square shaped movable piston so that the area of the square is just the side squared. OK. So we can finally solve for peace. So let's plug in all of our known variables. So the numerical value for the pressure of the atmosphere is 1. multiplied by 10 to the power of five pascals. And then we're given in the prom that the piston weighs kilograms multiplied by gravity. So the numerical value for gravity is 9.81 m per second squared, all divided by the area which we need to pause here because we're given the side as 15 centimeters, but we need to convert centimeters to meters. So we can use dimensional analysis to quickly do that. So 15.0 centimeters. So in one m there is centimeters and this is squared. So it's 38 kg multiplied by gravity 9.81 m per second squared divided by the area is squared. So when we plug that all into a calculator, the value that we should get for pressure is 1.178 multiplied by 10 to the power of five pass scales. Awesome. So now we can solve for V. So let's plug in all of our known variables to solve for V. So V equals the number of moles which in this case, it was 0.23 moles multiplied by the universal gas constant, which the numerical value for that is 8.31 joules per mole multiplied by Calvin. And then the temperature was in, was 300 Kelvin, all divided by the pressure which we determine the pressure value to be 1.178, multiplied by 10 to the fifth power pass scales. So when we plug that into a calculator, the value for V should be 0. m cubed. OK. So and finally to solve for L which is our end goal, let's recall and remember that volume is also equal to the area multiplied by the length. So we need to rearrange this equation to solve for L. So when we rearrange that using algebra, it b the length is equal to the volume divided by the area. OK. So let's plug in our known variables to solve for L. So we know that the volume we just found that was 0. 49 meters cubed divided by the area which was 15.0 C M. Let's do our quake dimensional analysis. Again. In one m, there is 100 centimeters squared. So when you plug that into a calculator, you should get 0.22 m, which is our final answer. Hooray, we did it. OK. So that means our final answer has to be B 0.22 m. Thank you so much for watching. Hopefully. That helped and I can't wait to see you in the next video. Bye.

The 50 kg circular piston shown in FIGURE P18.57 floats on 0.12 mol of compressed air. a. What is the piston height h if the temperature is 30°C?

Verified Solution

Key Concepts

Ideal Gas Law

Hydrostatic Pressure

Thermodynamic Principles