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0 Scc Hint Periodic Table 8See Hint803/17/20(0.3 point] atm (0.3 point) Part2 € ued Volume:(Graded)-WVin _ (0.3 pointh Part 1 Temperature: enment #7-

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The data in the following table are from a constantvolume gas thermometer experiment. The volume of the gas was kept constant, while the temperature was changed. The resulting pressure was measured. Plot the data on a pressure versus temperature diagram. Based on these data, estimate the value of absolute zero in Celsius. $$\begin{array}{rl}\hline T\left({ }^{\circ} \mathrm{C}\right) & P(\mathrm{atm}) \\0 & 1.00 \\20 & 1.07 \\100 & 1.37 \\-33 & 0.88 \\-196 & 0.28 \\\hline\end{array}$$

This problem were asked to complete this table shown up top. But before we can do that, we need to make sure everything is in the correct units. So I drum table again and copy down everything that's already in the right units, and we're gonna convert everything else. So for volume, we want our units to be leaders for temperature. We want Calvin's. And for pressure, we want atmospheres. So the View one column is already in leaders. So that's great. T one, uh, is all in Celsius. We need to be Calvin. So to convert from Celsius to Calvin, we're gonna add 273. So for 10 degrees Celsius, that gives us 2 83 Calvin zero gives those to 73 Calvin and 75 degrees Celsius. Gives us 348. Calvin R P. One column is all in atmosphere is already so. That's great. Um, t two, we have to convert from Celsius to Calvin again. So we'll have to 73 k 308 k and to 73. Okay. And then, uh, P two. We have one pressure in tour 1 80 Em is equal to 760 tour. So we will take that 7 35 and divide it by 7 16 And that will give us zero point 97 a t. M. Now to fill in the missing boxes in the table. We're going to use the combined gas law. So this tells us that p one times V one over t one equals p two times the two over t two. So for the first column here, r p one is 0.75 a t m b one is 6.35 leaders over T one of 2 83 Calvin, that is equal to 1.0 a t m for Pete to be, too, is what we're looking for in our teach you is to 73. Calvin. So to do this math first, we're gonna start with the left side. We're gonna most 5.75 by six point 35 and then divide it by 2 83 and we'll get 0.17 leaders a t m. Over Calvin equal to um one divided by 2 73 is Syria 730.37 a. T m. Over Calvin times be, too. Now we can divide that point. 017 by the 0.0, 37 And that will give us a B two of 4.6 leaders. That's what fills in that box for the second rover and do the exact same thing. So we'll set it up the same way. Just make some room. So P one is 1.8 e. M. Times V one is 75.6 leaders over to 73 K that is equal Teoh Syria 0.97 80 m times be to over the Rio eight k And if we solve that you doing the math the exact same way we'll get V two equal to 87.9 leaders and or last column we're gonna do the exact same thing. But this time will be solving for P two instead of be too. So, uh, on the left side we have 0.55 80 m for P one. B one is 1.6 leaders over 3 48 k for t one. We're looking for P two times 3.2 Leaders be to over to 73 k t two. So again we'll do the math the same away and we will get 0.14 80 m four p two.

In this problem, we're going to draw the phase diagram for are gone. So, based on the information provided to us, we can go ahead and label the phase diagram that I already went ahead and drew out. So on our excess axis we have temperature which I went ahead and put in degrees Celsius like given to us in the problem. And on the y axis we have pressure, which I would put in a. T. M. Given to us in the problem. And so looking at this diagram, we can kind of reason through the sectors, how they get labeled as we know at a higher temperature and a lower pressure. We would be in the gas phase as you can see over here and then in the as the temperature rises and the pressure stays relatively as the temperature remains high and the pressure starts to increase, we would see that we transition into the liquid phase and then as the temperature begins to decrease in the pressure stays pretty high, we go to the solid phase. Okay, so going ahead and labeling this, we can look at our normal melting point normal meaning it's at standard pressure. So standard pressure is 1 80 M. So we can go ahead and label that for the melting point. We'll put 1 80 M right here. We're going to go ahead and do the boiling point. So I'm just going to go ahead and extend that line. So this would be 1180 M. Sorry. one. Okay. And then for a boiling points we can labeled the temperatures so we'll go all the way down. So this would be our melting point and then this one is going to be our boiling point. And so they give us both, those numbers are melting point is going to be negative 187. Well on our normal boiling point is gonna be negative 186. And again, we're going based on sort of a standardized um diagrams already drew out, so this isn't completely drawn to scale. And so then they also go ahead and give us our triple point, which is right here where all the phases exist simultaneously. And so that will be At -189. So we can write that in and they also go ahead and give us the pressure that they put that at hoops. And so we'll go ahead and write that one into. So this would be negative one, squeezing that in there, we'll make this one over and get that under the line, oops, here we go. And then this would be a .6880 M. Okay. And then they also give us data for the critical point which is up here where the gas can exceed in normal circumstances, and so that one Would occur a negative one, and 48 8 p.m. So we can go ahead and write those in as well. And so after we've labeled everything, you can basically reason through any of the phase transitions that you would need for argon, and it will give you accurate detailing of the parameter surrounding those face changes. Thank you.

All right. In this problem, we're going to be using the table of values given to estimate the slope of the tangent line at P equals one P equals three. And because we're using the slope of a tanja line and estimating that we're going to use this slope of the Sikh in line All right, so that's f A B minus. Have have a all over B minus A. So at peak was one atmosphere. What we're going to do here is we're going to use the parts are the points around p equals one. So we're gonna use P equals zero and p equals two. And then we're also gonna use, um p of zero, um, or ff zero in this case, and that's going to be looking at the table. Um, we'll do that a second half of zero and f of two luscious, uh, but those there. So, um, now I plug everything into my formula of F A B minus f of a over B minus A. That's going to be equal to half of to minus F of 0/2 minus zero or more specifically again using the table negative 20 minus zero all over to minus zero. So this is gonna be negative. 10 negative. 20 over over to his negative 10 units per time. All right. And that's the estimate of the Tangela. And it's a good slope. All right. Estimates. So, what are the units? Will we have to kind of think about the units of what is R p values in and what are F of our function? Values and soapy is in, um, atmospheres. And, uh, me outputs are neg and degree Celsius. So this is going to be degrees Celsius over atmospheres because those units don't have any way of canceling. So we'll say, um, interpret. That's going to be, um, negative 10 degree Celsius per atmosphere. So that means that when the atmospheres of one, um, atmosphere, the temperature is decreasing at a rate of negative 10 C per each atmosphere. All right, now, a P equals three atmospheres. We're going to use P equals two and p equals four, which is then going to correspond to f of to and fro for all right. Um, so again, half a four minus f of to all over four, minus two. More specifically, that's going to be looking at the table negative 11 minus negative, 20. So that's going to be a plus 20 there all over four minutes to or just two. This is going to give me nine halves and again that's going to be in over degrees per atmosphere, and that means that's the rate of change at that point of time.

This problem assets to complete this table using the combined gas law. So the first thing we need to do is convert everything to the correct units. So this table of top here is what is given in the problem and then the table on the bottom. I'm gonna use to convert everything to the right units. So I've already copied down all the values that are in the correct units. Um, so we want our volume to be in. Leaders are temperature to be in Calvin's and our pressure to be in a T. M and atmospheres. So starting with the volume, we have one box here that is in mill leaders instead of leaders. One leader is equal to 1000 mill leaders, so mill leaders become 0.43 leaders, then in the temperature column to convert Celsius to Calvin. We have to add 273 to it. So 43 grief Celsius will become 316. Calvin negative 56 Celsius becomes to 17. Calvin and 25 degrees Celsius becomes to 98 Cullen then in the pressure column. We have 8 65 tour in the second row right here. So one a tm is equal to 760 tour. So we're gonna take that 865 and divide it by 760. And that is going to give us 1.148 c m. Um, and then the same conversion works with MILLIMETERS of Mercury one HTM is equal to 760 millimeters of mercury. So we'll take that 7 40 divided by 7 60 and we'll get 0.97 80 m here, Um, that is the same there. And then the temperature column again. We're just gonna add to 73 to everything. So we'll get 3 38 Kelvin's 3 16 Calvins, 30 to Calvin and to 73 coven. Now to solve for the missing boxes, we're gonna use the combined gas law which says that p one times the one over t one equals p two times me to over t two. So starting with this first row here, um, r p one is 6.5 a t. M. Times the one which is 546 leaders over t one is 3 16 Calvin that is equal to P two of 1.98 c m. Times B two over 3 38 Calvin So we can do that. Math 6.5 times 546 divided by 3 16 is 11.23 Leaders a T M. Over Calvin that is equal to one point nine divided by 3 38 k. So we get 0.56 a t m. Over k times the to. And if we divide 11.23 by that 0.56 we get that B two is equal. Teoh 1998 Leaders There is that box right there. For the second row, we have a P one of 1.148 c m. The one of 0.43 leaders over 2 17 carry equal to make some room here equal Teoh P two of 1.5 80 m times the two which we're looking for over 3 16 Calvin. So if we do that math the exact same way we did before, we will get a zero point 048 leaders as our answer there in the third row. We have a P one of 0.87 a. T. M. Times 4.2 Leaders over 2 34 k Equal Teoh Now we're looking for P two times 3.2 leaders over 302 k Again that math will do it the exact same way we did the 1st 1 and we will get P two equal Teoh 1.47 a. T m. And for the last column, you make some room again. We'll start with people on a 0.97 a. T. M. Times 1.3 leaders over to 98. Calvin It's equal Teoh one a. T m. Times B two over to 73 k again will solve for that same ways we did before and we'll get 1.15 leaders or be to there.


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