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

b. How far does the piston move if the temperature is increased by 100°C?

Question

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

b. How far does the piston move if the temperature is increased by 100°C?

Accepted Answer

Hello, fellow physicists today, we're gonna 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. A circular piston could vertically move due to the expansion or compression of 0.15 moles of compressed nitrogen gas. As shown below. What is the change in position in meters of the piston? If the nitrogen temperature is increased from 700.0 degrees Celsius to 808 100.0 degrees Celsius. OK. So we're given some multiple choice answers and they're all in the same units and meters. Let's read them off to see what our final answer might be. A is 6.0 multiplied by 10 to the power of negative four. B is 4.9 multiplied by 10 to the power of negative three C is 2.2 multiplied by 10 to the power of negative two and D is 5.5 multiplied by 10 to the power of negative two. OK. So, well, first off, let's make the following assumptions. Nitrogen behaves like an ideal gas. The pressure of the compressed nitrogen is constant. The number of nitrogen molecules is constant. The change in the height of the piston is directly proportional to the change in the gas temperature. And that the system is in a is in static equilibrium when the gas temperature is at 700.0 degrees Celsius and at 800.0 degrees Celsius. So our first step is to determine what the pressure is. So please note that this is an isobaric process. So that means that the pressure remains constant. We also need to recall 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. We also need to make a quick note that the volume increases due to the temperature increasing. OK. We also need to note that the pressure is equal to the amount of the total pressure exerted by the atmosphere and the weight of the piston compressing on the nitrogen gas. So we can recall and write that the pressure is equal to one atmosphere plus the weight divided by the area for the piston. OK. Subscript piston. OK. So first off, let's start by solving for the pressure and then we can solve for the volume. So let's solve for the pressure. First, let's plug in all of our known values to solve for that. So the numerical value for the pressure of the atmosphere is 101.3 multiplied by 10 to the power of three past scales plus. OK. So before we move on, let's make a quick note here that the pistons mass is 50 kg as stated in the diagram right here. So we need to use that and we need to multiply it by the gravity because as we shouldn't remember that weight is equal to mass multiplied by gravity. So as stated in the diagram, the piston is 50 kg and we need to multiply it by gravity which is 9.8 m per second squared. And that's all divided by the area which we need to recall also that an area of a circle states that pi multiplied by radius squared. So that's the area of a circle, but we're given the diameter instead. So we need to note that radius equals diameter divided by two. So taking that into account, define the area of the circle. In this case, you need to take the centimeter value and convert it to meters which when we convert 15 centimeters to meters, it's 0.15 m and we divide it by two and that value is squared. OK. So then when we take all that and we plug it into a calculator, we should get 129. multiplied by 10 to the power of three past scales. Now we can solve for the volume. So if we take the ideal gas law equation, let's call it equation one and we rearrange it to solve for V, it will look like this. So delta V, so the change in V equals the number of moles, number of moles multiplied by the universal gas constant multiplied by the change in temperature. Delta T all divided by the pressure. OK. So now at this stage, you can plug in all of our known variables to solve for V. So let's do that. So that when the problem we're given that there's 0.1415 moles, so 0.15 moles multiplied by the universal gas constant, which the numerical value for that is 8. and its units are jewels per Kelvin multiplied by mole multiplied by the change in temperature. But let's take a minute here in pause because the temperatures given to us in degrees Celsius, we need to convert the degrees Kelvin. So let's do that really quick. Let's remember how to do our conversion factors. So T I, the initial temperature and then we need to find the final temperature. So let's start with the initial temperature. So it's so all we have to do to convert is we take the initial temperature which is 700 degrees Celsius and you just add 273.15 which equals 973. Calvin for the initial temperature. And we do the same thing for the final temperature which is 800 degrees Celsius. And we just add 273.15 which equals 1073.15. Kelvin. So to do the change in temperature, we take the final temperature minus the initial temperature. So 1073.15 minus 973.15. And those are both Calvin. Don't forget our units here all divided by the pressure which we determine to be 129.0, multiplied by 10 to the power three past scales. So when you plug that into a calculator, the change in volumes, the volume should be 9.7 multiplied by 10 to the power of negative four m cut. And that's the simplified version when you write it as a sign. When you write it in scientific notation, if you notice when you type it into your calculator right off the bat, it will look something like this. 0.967 m cubed. But if you convert that to scientific notation, it makes it quicker to type in your calculator. OK. So now we can solve for our final answer hurray. So we need to find the change in position. So in order to find the change in position, we need to use the equation for the volume of a circle. So let's recall that the volume of a circle equal is V equals pi R squared multiplied by the height. But in this case, we're trying to find the change of position which as indicated in the diagram, it's gonna be delta Y. So rewriting this, considering that we're trying to solve for delta Y, which is the change in position. And also that we're given it as diameter set of radius. The equation will look as false. So the volume of the circle is equal to pi multiplied by diameter divided by two squared multiplied by delta Y which is what we're trying to find. So now we need to rearrange the volume of a circle to solve for delta Y which is the change in position. So when we do that, the equation should look like this delta Y equals four multiplied by delta V divided by pi multiplied by diameter square. So now we can plug in our numerical values to solve for Y. So let's do that. So four multiplied by our volume which we determine to be 9.7 multiplied by 10 to the minus four meters cubed all divided by pi by the diameter squared which was 0.1 five m squared. So when you plug that into a calculator, your final answer will look something like this. 0.549, OK,  aN:aN:000NaN 0. m. Which when you convert that the scientific notation really quick will be 5.5 multiplied by 10 to the negative two m and that is our final answer. OK. So just really quick, just in case you're confused, we're given in our diagram that the diameter is in centimeters. So had to convert that to meters just to be clear. OK? So that means that our final answer is D 5.5 multiplied by 10 to the power of negative two 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. b. How far does the piston move if the temperature is increased by 100°C?

Verified Solution

Key Concepts

Ideal Gas Law

Thermal Expansion

Pressure-Volume Relationship