A constant-volume gas thermometer registers an absolute pressure ...

Question

A constant-volume gas thermometer registers an absolute pressure corresponding to 325 mm of mercury when in contact with water at the triple point. What pressure does it read when in contact with water at the normal boiling point?

Accepted Answer

Hey everyone welcome back in this problem. We have fixed volume containers being used as thermometers. They were told the temperature of a sample is reported in absolute pressure And were asked if the thermometer displays an absolute pressure of 420. Mercury at the triple point of carbon dioxide determined the pressure displayed at the boiling point of Ethanol. Mhm. Alright. So what we wanna do is we want to find a pressure at a particular temperature. We have pressure and temperature and we're talking volume and we think of the ideal gas law or the gas equation. Okay, we can write this as PV equals N. R. T. Okay. All right, so let's think about this. Okay, we're gonna have two versions of this equation. Ok, we're gonna have the version um where we have the initial Okay, pressure, initial volume, an initial temperature and then we're gonna have the pressure volume and temperature final at this boiling point of ethanol. That's what we're looking for. Well, we're told that these are fixed volume containers. Okay, so this V. I. Is actually going to equal VF. Okay, so we can get rid of the subscript here. Those are the same. Okay, we also have no matter exchange here. Okay, this is closed. So N. R. Is also gonna be the same in both equations. Okay, So if we rearrange, we can rate P. I over T. I. Is equal to N. R Over V. Okay, Which is just going to be a constant. Similarly over here, we can write PF over T. F. Is going to be N. R over V. And this is going to be equal to the same constant K. Because these values are fixed constants in this problem. Okay, Since these are equal to the same constant, we can set them equal to each other. P. I. Over T. I. Is equal to P. F. Over TF. Okay. And this is sometimes called the gay lucic's law. Okay? Um and so depending on the textbook, depending on the professor, you can jump right to this if you know you're in a fixed volume situation where you're not exchanging matter. Okay? Alright. So filling in the information, we know the initial pressure we're told is of mercury. We have 420 of mercury. Our initial temperature is at the triple point of carbon dioxide. The triple point of carbon dioxide which you can look up in a table in your textbook. Okay. is negative 57°C. And We're Gonna Add 273 groups 273, not 275. So that we can write this in terms of kelvin on the right hand side, we have P. F. That final pressure. And that's what we're looking for. And we want to know that final pressure when we're at the boiling point of ethanol. And again, looking this up in a table in your textbook, the boiling point of Ethanol is 78°C. Okay. And we're gonna add 273 in order to write this in terms of Calvin. All right, ourselves a little bit more room here and we can rearrange and we get that final pressure is equal to 420 of Mercury times 351 Calvin, divided by 216 Calvin. The unit of Calvin will cancel. We're going to be left with mm of mercury and that is the final pressure at the boiling point of Ethanol that we were looking for. This is going to correspond with answer d. That's it for this one. Thanks everyone for watching. See you in the next video.

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