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Question

In Problems $67, 68, 69,$ and $70$ you are given the equation(s) used to solve a problem. For each of these, you are to draw a pV diagram.

$(T_2 + 273) ext{ K} = \frac{200 ext{ kPa}}{500 ext{ kPa}} imes 1 imes (400 + 273) ext{ K}$

Accepted Answer

Hey, everyone. Let's go through this practice problem. Construct a pressure versus volume diagram for the equation given below V sub two equals 550 plus 273. Kelvins divided by 120 plus 2 73. Kelvins multiplied by one multiplied by 350 cubic centimeters. All right. Now, before we get to drawing the diagram that we're asked to construct, I first want to address this equation because it might look confusing at first and you might not be entirely sure what this means or why it's set up the way it is. So to illuminate that let's start by using the combined ideal gas law which states that for a given transformation P one multiplied by PV, divided by T one is equal to P sub two multiplied by V sub two divided by T sub two. Now, this equation is an equation for V sub two. So let's rewrite this algebraically for V sub two by multiplying both sides of the equation by T sub two and dividing both sides of the equation by P sub two. And what we find is that V sub two is equal to T sub two divided by T sub one multiplied by P, sub one divided by P, sub two multiplied by V sub one. Now we can compare the form of this equation to the equation that the problem gave us. And now it's a little more clear what's happening and what the different variables represent. The temperature in the denominator converted into Kelvins is T sub one, the initial temperature of the system. So T sub one is equal to 100 and 20 degrees Celsius. Similarly, the temperature in the numerator converted into Kelvins is T sub two, the temperature after the process has taken place. So T sub two is equal to 550 degrees Celsius. The volume shown on the right hand side of the equation given is V sub one, the initial volume or 350 cubic centimeters. The equation given doesn't have any terms that look like pressure, but we are given very specifically a one term in the equation which seems to indicate that P sub one divided by P sub two are equal to one implying that the two pressures are equal. So P one is equal to P two, the pressure in the process never changes. And so now that we have all that information, let's try constructing the pressure versus volume diagram. So here's the axis. Let me try that again. Here is the setup and this vertical axis is going to represent pressure and the horizontal axis is going to represent a volume. And I will say the volume has units of cubic centimeters first off. Since we know that the pressures in both sides of the process are the same. That means that whatever graph we draw, it's just going to look like a horizontal line since pressure isn't going to change at all. However, in order to determine where the line starts and stops, we first need to know what the second volume actually is. So let's actually take the equation that the problem gives us and just plug it into a calculator to find what V sub two is. That way, we'll know what the final volume it is. So just plugging the equation given by the problem into our calculator without changing anything up, and we find that the final volume V sub two is equal to about 732. cubic centimeters. So now we have the initial and final volume and the final volume is 732.952 which means that our highest axis, our highest point that we label on the volume axis of our diagram doesn't need to be much higher than that. So I'm going to cap the volume axis at 800 cubic centimeters, which means that the halfway point is gonna be 400 then the intermediate terms, the intermediate labels are gonna be divided accordingly. So from 0 to 4, that's gonna be like 1 to 3 and then from 4 to 8, 100 it's gonna be 506 107 100 right. So now we're ready to begin labeling the plot. So the initial volume of the sub one is 350 cubic centimeters. So it's going to start right in between the 304 tallies. So it's going to start somewhere right here and I'm using a dotted line to represent where it goes. And because we don't know anything about the pressure, but we know the pressures are equal. We don't need to worry about labeling the pressure axis. We can just kind of choose an arbitrary height at which to draw the process. And then the final volume V two is 732.952 cubic centimeters. So it's going to be sort of in between the 708 100 tallies except a little bit to the left of the middle since it's underneath 750. So it's gonna come up somewhere here. So our initial point is going to be at the top of the left most dotted line because that's the point where the temperature is 100 degrees Celsius. And the right most position on the right dotted line is going to be where the temperature reaches degrees Celsius. And since the left, most point corresponds to variables one and the right most point corresponds to state two, that means that I'm going to draw an arrow pointing from 0.1 to 0.2. So I'm gonna have a little arrow head showing the direction of the process. And really, that's all we needed to do. We have fully drawn the diagram. So that is it for this problem? I hope this video helped you out. If you think you need more practice, please check out some of our other videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

In Problems $67, 68, 69,$ and $70$ you are given the equation(s) used to solve a problem. For each of these, you are to draw a pV diagram.
$(T_2 + 273) \text{ K} = \frac{200 \text{ kPa}}{500 \text{ kPa}} \times 1 \times (400 + 273) \text{ K}$

Key Concepts

pV Diagram

Ideal Gas Law

Temperature Conversion

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