A container of gas at 2.0 atm pressure and 127°C is compressed at constant temperature until the volume is halved. It is then further compressed at constant pressure until the volume is halved again. Show this process on a pV diagram.

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Hey, everyone. Let's go through this practice problem. A gas in a piston has an initial volume of 500 mL, a pressure of 1.5 atmospheres and a temperature of 25 degrees Celsius. The gas then undergoes isothermal compression, decreasing its volume to 250 mL. Next, the gas is cooled at a constant volume until the pressure is reduced to one atmosphere. Draw this two step process on a pressure volume diagram. OK. So we're drawing a PV diagram of a two step process. And what that basically means is that there are going to be three dots and two line segments on our graph where each dot represents a state that the gas is in and each line segment will represent one of the steps of the process. So if our gas is in an in in an initial state of state, one, then the first step of the process will transform the gas into its second state where some of the variables have changed. The second step of the process will turn it into its third. And in the case of this problem in the final state, in order to represent the process on a PE V diagram, we need to know both the pressure and the volume for all three states of the process. Unfortunately, for us, the problem already gives us most of that. The problem tells us the initial volume and the initial pressure. So we have P one and V one. Then when the first process happens, we're told the second volume V two, but we aren't told P two. And then when the final step of the process happens, we're told the final pressure of the gas. So P three and we're not directly told the final volume V three, but we're told that the volume is constant during the second step of the process. So we pretty much know V three as well just by the virtue of knowing V two. But the only thing that we need to do actual math to find is P two, the second pressure, the pressure after the first step of the process. And since we know all the other variables related to the gas, at that point, we confine to this, using the ideal gas law to compare the variables between two states of the problem. For this case, I'll use the ideal gas law to compare the first two states states one and two. So according to the combined form of the ideal gas law, the pressure multiplied by the volume divided by the temperature is always constant. So for our purposes, P sub one, V, sub one divided by T sub one is equal to P sub two V, sub two divided by T sub two, we can simplify this a bit because the problem tells us that for the first step of the process, there is an isothermal compression which means that temperature remains constant. So that means that T sub one is equal to T sub two. So the TS can cancel out of our expression. So we're left with P sub one V sub one is equal to P sub two V sub two. So remember we want to solve for P sub two. So we can algebraically find a formula for P sub two by dividing both sides of the equation by V sub two. So that tells us that P sub two is equal to the initial pressure multiplied by the initial volume divided by the final volume. So that's the initial pressure of 1.5 atmospheres multiplied by the initial volume of 500 mL all divided by the final volume of 250 millilitres. We don't need to do any unit conversions on the volumes because the milliliters will cancel out. And what we find by putting this into a calculator or just doing the calculation by hand that the final pressure is equal to 3.0 atmospheres. So this is P sub two. And now that we know the pressure and volume at all three states of the problem, we we are now ready to begin drawing the diagram. So the first step of course is going to be setting up her axes. So there's going to be a horizontal volume axis with units of milliliters. And there's going to be a vertical pressure access with units of atmospheres. I'm choosing these units for pressure and volume because these are the units that are given to us in the problem. So it makes things convenient. The next step is to set up the scale. Of course, I'm going to assume the origin has a zero for both the pressure and the volume. And the way I like to set up the scales is to find the maximum value for each variable and then base the scales around that. So if we look at the information given to us in the problem and at the information that we've discovered the things we've solved for the highest volume that we'll need to worry about is 500 mL. So I am going to set the high point on the volume scale to 500 then I'm just gonna kind of divide it up based on that. So about halfway through is gonna be where 250 mL is and so on and so forth. Now, for the pressure of all the pressures given to us in the problem, the highest one that we have to deal with is the one we found for state two, the three atmospheres. So I'm gonna set the high point on the pressure scale, the 3.0. And then once again, I'm going to divide the scale up accordingly. So about halfway down is where 1.5 atmospheres is. Now let's begin labeling our points. So the problem tells us that the gas's initial volume is 500 mL and the initial pressure is 1.5 atmospheres. So that means our first point will be on the 500 mL side of the volume scale. And it's going to be on the 1.5 atmosphere side of the pressure as I'm representing with the dotted lines. So our first point is going to be right where the dotted lines intersect right here. And this is 0.1 for the second state. The problem tells us that the volume is decreased to 250 mL. So the next volume point is gonna be on the somewhere on the 250 mL volume axis, but with an isothermal compression and we also found the new pressure to be at three atmospheres. So 250 mL and then up to air, we have three atmospheres of pressure. And so that's going to be where the second dotted line is the the second dots for state two, though we also have to draw a line connecting state one and two to show the flow of the process. And remember the problem tells us that there is an isothermal compression. So the temperature is constant. And as we discussed earlier. If the temper, if it's an isothermal process, then PV, the pressure multiplied by the volume is constant, which means that the curve uh uh the the line connecting them is going to have kind of an inwardly curved shape because that's how isothermal curves look like actually make that little meter. So kind of like that with an arrow showing the flow of the process. And finally, the gas is cooled at a constant volume until the pressure is reduced to one atmosphere. So the final pressure is going to be one atmosphere. So I have not labeled that on my scale yet, but kind of dividing this into thirds. Here's gonna be where 0.5 atmosphere fears are. Here's gonna be where one atmosphere is. So it's gonna be right about here. And since the problem tells us that it's a constant volume, the volume doesn't change. So we're remaining on the same volume axis of 250 mL. And this is where the final point point three is. And since, and since the volume remains constant throughout the entire step, there is just a straight line connecting the two of them. So from 2 to 3, you have an arrow pointing downwards. And that is our answer to the problem. We don't have any multiple choice options to choose from. But our logic is sound and really, that's it for this problem. I hope this video helped you out if you like more practice. Please check out some of our other videos which will give you more experience with these types of problems. That's all for now. I hope you all have a lovely day. Goodbye.

A container of gas at 2.0 atm pressure and 127°C is compressed at constant temperature until the volume is halved. It is then further compressed at constant pressure until the volume is halved again. Show this process on a pV diagram.

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Key Concepts

Ideal Gas Law

Isothermal Process

pV Diagram