Suppose 3.0 mol of neon (a monatomic gas, assume ideal) at STP...

Question

Suppose 3.0 mol of neon (a monatomic gas, assume ideal) at STP are compressed slowly and isothermally to 0.22 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume. Find the highest and lowest temperatures and pressures attained by the gas, and show on a PV diagram where these values occur.

Accepted Answer

Hey, everyone in this problem, we're asked to imagine that we have a container with an ideal mono atomic helium gas quantity equal to four moles at standard temperature and pressure. The helium gets compressed slowly in an isothermal manner until it reaches 1/5 of its original volume. Then allowed to expand rapidly in an adio manner. Back to its original size, were asked to determine the highest and lowest temperature and pressures attained by the helium and show this on a PV diagram where the values occur. And we're told to use the universal gas constant R 0.0821 L atmospheres per Kelvin. So let's get started. We have a blank diagram here. We have our volume and leaders on the X axis, our pressure and atmospheres on the Y axis. And we're gonna come back to this and plot this diagram. Oh We're gonna start by writing out this information that we've been given. So let's start with this first time point. OK. So initially what we have is that the temperature is at standard temperature and we have standard temperature and pressure. So our temperature which we'll call t one that initial temperature is gonna be our standard temperature 273 Kelvin and the pressure P one will be our standard pressure one atmosphere. Yeah, we're gonna leave the pressure in atmospheres because that is what is on our diagram, on the Y axis, we'll leave it in that unit. OK. Now, we also know that we have four moles and that's not gonna change through these time points. So we know that N will be equal to formal and we don't know yet what the volume is at this. OK? So that's the first point. Then we're gonna undergo a process and we're told that this is isothermal, we undergo this isothermal process until we reach 1/5 of the original volume. Now, because this is an isothermal process, the temperature is gonna remain constant. So the temperature at time 0.2 is still gonna be 273. Kelvin, we don't know what the pressure is going to be. And the volume V two, we know will be equal to 1/5 of the volume V one. Then we're gonna go through another process and this time we're expanding in an adiabatic manner. So we have this audiomatic expansion and that is gonna take us to this third time point where we have some temperature, some pressure and we're going back to our original volume. So V three will be equal to V one. OK. So what we wanna do is find the highest pressures and volumes and indicate those on our graphs as well as our temperature. OK. So we need to find all of these values. And we're gonna start at this first time point, we already know the temperature and the pressure, we know the number of moles we can use our ideal gas law. So for time 0.1 we have P one V one is equal to N RT one. OK. Again, by the ideal gas law. So the missing piece right now is that volume V one, we can rearrange and write this as V one is equal to N RT one divided by P. What substituting in our values V one is equal to formal multiplied by 0.0821 leader atmosphere. Alvin multiplied by 273 Calvin divided by one atmosphere. Now the unit of mole will divide out, the unit of carbon will divide out, the unit of atmosphere will divide it and we're gonna be left with liters, which is what we want for volume. So this works out great. We get that our volume V one will be equal to 89.6532 L. OK? So that's the first volume. Now we have that volume. So we know that the volume V two which will be 1/5 of the initial volume, you know that V two will be 1/5 of 89.6532 L, which is equivalent to about 17.93064 L. OK. So we have our volume V two. Let's now find our pressure and we're gonna do it very much in the same way. We're gonna use the ideal gas law again. So if we go back to our ideal gas law, we can write P two, V two is equal to N RT two. We wanna solve four P two. So P two will be equal to N RT two divided by V two. Substituting in our values P two is equal to formal multiplied by 0.0821 L atmospheres for Calvin mole multiplied by 273 Calvin all divided by 17.93064 L. And we get a pressure P two that is equal to five atmospheres. OK. So let's go back up and take a look at our information. Where are we at so far? Well, so far we've figured out all the information for this first time point. So we have our volume. Now we have all of the information for the second time point. We have the pressure, we just calculated, we have our going and now we can move on to that third time point. And for the third time point, we're missing both the temperature and the pressure. So that makes it difficult to use that ideal gas law. What we're gonna do is make use of the fact that this is an adiabatic process. OK. In an adiabatic process, recall that we have a relationship between the pressure and the volume. So we can say that the pressure P two multiplied by the volume B two to the exponent gamma is equal to the pressure P three multiplied by the volume B three to the exponent gamma where gamma is the ratio of heat capacity. OK? We know the volume, we know P two. And so this will allow us to calculate that pressure P three. Then we can use the ideal gas law to calculate T three. So let's start with finding this value of gamma. So we're called gamma. The ratio of heat capacities is going to be CP divided by CB. So the heat capacity at constant pressure divided by the heat capacity at constant volume. Now we're told we have a mono atomic gas because we have a mono atomic gas, that means CV will be equal to three halves multiplied by R. And because we have an ideal gas, we can say that CP will be equal to CV plus R which in this case will be five hr which tells us that this value of gamma will be five halves are divided by three halves are, which gives us a value of five thirds. OK. So we have this value of gamma. If we work off of our equation for the adiabatic process, we wanna solve for the pressure P three. So we can write that P three will be equal to P two V two to the exponent gamma divided by V three to the exponent gamma. Substituting in our values P three will be equal to five atmospheres multiplied by 1/5 of the one to the exponent gamma divided by the one to the exponent gamma. So instead of writing out the value for V two and V three, what we've done is written them both in terms of V one. What that allows us to do is cancel out that V one to the exponent gamma, we can save some round off error and we can simplify this equation without even having to substitute those values in. So we get that our pressure P three will simplify down to five atmospheres multiplied by 1/5 to the exponent gamma. So to the exponent five thirds A V one to the exponent gamma will divide out and we are left with this pressure P three being equal 25 atmospheres, 1/5 to the 15 3rd, 0.341995 atmospheres. And we have one step left, we can fill in our pressure. In our information. We have one step left to find this temperature and then compare what we found to find these highest and lowest values. So going back to our ideal gas law and rearranging, we can write that the temperature T three will be equal to P three V three divided by nr substituting in our values T three will be equal to 0.341995 atmospheres multiplied by 89.6532 L divided by four more multiplied by 0.0821 L atmosphere per Calvin mole. And you will also often see these equations in terms of jewels and pascals, those sort of units. In this case, it's OK because these units are gonna divide it. So what we need to make sure is that all of our units are consistent. So the ideal gas constant that we use needs to have the correct units for the units we're dealing with. And in this case, we're OK. So those atmospheres will divide it, the leaders divide it, the malls divide it. We're gonna be left with Kelvin, which is what we want for temperature. When we get that T three is going to be 93 0.364 635. So if we think about our highest temperature and our highest pressure, let's go back to all of this information we wrote out and take a look at what we have so that we can compare it all. Let's write in that final temperature we just found. And what we can see is that the highest temperature we have is 273 Kelvin, which occurs at time 0.1 and two. The highest pressure we have is five atmospheres which occurs at time 0.2 the lowest temperature we have is T three and the lowest pressure we had is P three. If we take a look at our graph and we plot these points that we have, we have our temperature T one. So we have our pressure of one atmosphere corresponding with a volume of 89 right around here. OK. So that's our first time point. And we have our second time point, we're gonna increase the pressure and our volume is going to decrease. I'm gonna be right around here and then we have our third time point where we have the same volume as initially, but we have a lower pressure. So that's gonna be right about there. And we can see that that green dot that's our lowest pressure and we can see that our lowest volume will be our red dot ok. So those are those points plotted on that graph and that's it for this problem. We've identified those high and low points. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Ideal Gas Law

Isothermal Process

Adiabatic Process

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