(II) A 1.0-L volume of air initially at 3.5 atm of (gauge)pres...

Question

(II) A 1.0-L volume of air initially at 3.5 atm of (gauge)pressure is allowed to expand isothermally until the (gauge) pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume.

(b) How much work does the 1.0 L of air do in this process?

(b) How much work does the 1.0 L of air do in this process?

Accepted Answer

Everyone in this problem, we're asked to calculate how much work is done by 2 L of helium gas. When it's initially at a pressure of five atmospheres, the gauge pressure and then the gas undergoes a three step process as follows. Step one, the gas is allowed to expand isotherm until the pressure drops to one point that atmosphere two, the gas is then compressed at constant pressure back to its initial volume. And finally, the gas is heated at constant volume until it reaches its original pressure. We're given four answer choices all in atmosphere liters. Option A 1.5 option B 3.5 option C 5.0 and option D 7.0. So let's start by writing out what we have going on here. So initially, we have a volume vnat of 2 L. We have a pressure peau well, we have a gauge pressure of five atmospheres to get our absolute pressure. We need to add the atmospheric pressure to that gauge pressure. The atmospheric pressure is one atmosphere. So that's gonna give us a pressure of six atmospheres and we don't know the temperature. And so that's the initial state of what we have then we undergo this process process. Number one, that is isothermal. So we undergo process number, process number one brings us to some volume vy it brings us to a pressure P one of 1.5 atmospheres. Because the first pressure we were given is gauge pressure. We're gonna assume that all of the pressures we are given are gauge pressures. So 1.5 atmospheres, we have to add that one atmosphere in order to get the absolute pressure, which is gonna be 2.5 atmospheres. And we know that the temperature T one must be equal to the temperature TN. And because this is an isothermal process, the temperature is constant, now we go to the next phase and the next phase is this second process. In this process, we're told has constant pressure. So because it has constant pressure, the pressure at this point is gonna be the same as the pressure at the time point before it. Now we're told it goes back to its initial volume. So the volume V two will be equal to 2 L. That initial volume, the pressure P two, as we just mentioned will be equal to P 12.5 atmospheres. And we don't know what the temperature will be. And we have our final process and our final process we're told is at constant volume. That's gonna take us to a point where we have a volume V three of 2 L if the volume is constant, it doesn't change when we come back to the original pressure. So P three is going to be six atmospheres and we don't know what that final temperature will be. OK. So we have three processes here. And what we're gonna do is go through each of, we wanna find how much work is done and the total work done w total will be equal to the work done by the first process, we lost the work done by the second process. We lost the work done by the third process. And we are gonna go ahead and calculate those all individually and then add them together at the end. So let's start with this first process, this isothermal process. OK? Now, when we're undergoing an isothermal process, we're called it the work done. In this case, we call W I, it is gonna be equal to N the number of MTs multiplied by R the gas constant multiplied by T the temperature multiplied by lawn of the final volume divided by the initial volume. Yeah, what we have is the volume V I, we have the initial volume. We don't have the volume we want. So we don't have the final volume and we don't have the temperature either. So let's write some other things out and see what we can do here. Now recall we have the ideal gas law when the ideal gas law tells us that PV is equal to N RT. So in this case, in our isothermal process, we can write the ideal gas law for the two different points. So we have PNVN is equal to N RT not and we have P one V one is equal to N RT one. Now remember what we know about the isothermal process, the temperature remains the same. So T knot is equal to T one and we can call both of these things tea, we just have one temperature. So if we think about this, in terms of our equations, we can write PNVN is equal to N RT and we can write P one V one is equal to N RT. OK. The right hand side of these equations is the same and RT are all the same for both equations. So the right hand side of the equations is the same. That means the left hand side must be the same as well. So we must have PNV not equal to P one V one. What this is gonna do is allow us to calculate that missing volume V one. We weren't able to do it directly off of the ideal gas law because we didn't have the temperature. But now that we've narrowed it down, we have an equation with the two pressures that we know the initial volume we know we can solve for V one. So let's rearrange V one will be equal to peanut V, not divided by P one substituting in our values. V one will be equal to six atmospheres multiplied by two leaders divided by 2.5 atmospheres. And this gives us a volume V one 4.8 L. OK. So now we have that final volume, we have the initial volume. What about this N RT value? OK. I'm not sure that we were given the number of moles. We still don't know that temperature. T however, remember that we wrote our equation P not VNA is equal to N RT. So we can replace the N RT in our work equation with P not Vena. OK. So we can write that the work W I is equal to PNVN multiplied by lawn A V one divided by V. Now, we could have also written this as P one V one as well. Um Either one is works because we know both of them are equal to N RT. OK. So we could have done either of the two, we just chose to go with that initial. OK. Now we have this equation where we know all of the values in it, we need to substitute them in and solve. So the work for this first process is going to be equal to six atmospheres multiplied by 2 L multiplied by law of 4.8 L divided by 2 L. And we get that this work for process one, about 10.5056 atmosphere liters. And you can see that we didn't convert our pressure into pascals, we didn't convert our volumes into meters cube. And that's because of the unit, the answer choices were given. OK? So these units match the answer choice. So we have this first part of our work. Now we're gonna move on to this second process. OK. Let's go back up. The second process is a constant pressure process. OK? So we're on to process two and we have constant pressure. What that means is this is an isobaric process. If we have an isobaric process, then recall that the work done is just equal to the pressure multiplied by the change in volume. All right. Well, that's gonna be equal to the pressure P one. Yeah, or P two, they're the same. So either way multiplied by the change in volume V two minus V one substituting in our values, the work done in this process is gonna be equal to 2.5 atmospheres multiplied by 2 L minus 4.8 L. And we get a negative value here. We get negative seven atmosphere liters and be careful with these signs. Be careful with terms like the change in volume, make sure you're writing them in the correct an in the correct order. It is possible to have negative work and we need to account for that in that total work done. All right. So we're done with process two. We're moving to process three and process three is a constant volume process. OK? And because this is constant volume recall that the work done is just going to be equal to zero. OK. So the work done in a constant volume process is zero. So our work done in this case is zero and that makes it simple. We're already done with that process and we can move on to our final answer. So our total work it is gonna be the sum of all of these works we found. So W 1 W-2 and W three. So the total is going to be the one W-2 W three, all three of those values we've already found substituting them in we have 10.5056 atmosphere leaders alas negative seven atmosphere leaders. Who was the OK. So those are the three values we just found. And when we add those up, we get a total work that is equal to about 3.5056 atmosphere liters. And that is the answer we were looking for. That is a total work done throughout this entire process. Going back to our answer choices and comparing what we found. We're going round to two significant digits and we can see that the correct answer is going to be option B 3.5 atmosphere liters. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Isothermal Process

Work Done by a Gas

Thermodynamic Cycles

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