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Problem %. 20 pts. Suppose we prepare |.,397kg of henzene S0 that it Is exaetly its melting/freezing point. bur still entirely liquid, and we immediately place it i...

Question

Problem %. 20 pts. Suppose we prepare |.,397kg of henzene S0 that it Is exaetly its melting/freezing point. bur still entirely liquid, and we immediately place it in a refrigirator: Suppose the refrigerator Ikes away hent frm the henzene Neady rale P . Suppose further (hat it takes 2960 completely treeze the benzene Iromn the moment is placed into the refrigeralor; What fs F'! (Don"forgetto include units') The latent hent f melting O benzene fs 4.27x 16'J/kg:

Problem %. 20 pts. Suppose we prepare |.,397kg of henzene S0 that it Is exaetly its melting/freezing point. bur still entirely liquid, and we immediately place it in a refrigirator: Suppose the refrigerator Ikes away hent frm the henzene Neady rale P . Suppose further (hat it takes 2960 completely treeze the benzene Iromn the moment is placed into the refrigeralor; What fs F'! (Don"forgetto include units') The latent hent f melting O benzene fs 4.27x 16'J/kg:



Answers

The melting point of benzene is $5.5^{\circ} \mathrm{C}$ and its boiling point is $80.1^{\circ} \mathrm{C} .$ Sketch a heating curve for benzene from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ a. What is the state of benzene at $15^{\circ} \mathrm{C} ?$ b. What happens on the curve at $5.5^{\circ} \mathrm{C} ?$ c. What is the state of benzene at $63^{\circ} \mathrm{C} ?$ d. What is the state of benzene at $98^{\circ} \mathrm{C} ?$ e. At what temperature will both liquid and gas be present?

Problem. 88 were asked to draw a heating grouper Benzie. Okay, so we're given the melting point boiling point. And we're doing a melting curve or heating girl from zero degrees sauces to 100 degrees sauces, which I have, um, pretty labeled, um, Intergraph on the right. Okay, so So what I want to do here is take that melting point and just trace it in our graph. OK, so 5.5 would be right around here in the middle of zero and 10 so I just want to trace that as such. Good. Next boiling point would be 80.1, which is right, right around that 80 mark. Okay, so just trace that as well. And I'm doing this because this is where our horizontal lines were going to be. Once we draw that heating group, so is to go for it. Um, as usual, we start and, um, solid state until we hit that melting point, which is 5.5. Okay, So once we hit that to going to be a horizontal line, and then we heat that liquid becomes liquid at this point, and that just goes, the temperature just goes up until you hit the boiling point. Okay, so at this point, it's going to be another horizontal line. This is where you are boiling, and then we increase the temperature all the way to 100. Ask. Spare the instruction scape. Try to draw that. Okay, so we hit 100 were good. All right, so that would be our, um, heating group. So what is the state of 15 degrees sauces at 15? Um, this is where 15 degrees sauces is between seven. And 20. And if you trace that, you hit the part of the curve where you have liquid, so that would be liquid. Next, Um, at 5.5 degrees Celsius. We saw that. Because it is it is the multi point. Okay, you're you're curve would be a flat horizontal line, so just right. Flat horizontal. Okay. Next, at 63 degrees Celsius right around here. Um, we traced that all the way to our graph and received that we hit the part of a cover. It's liquid. So that would be your liquid state. Last one at 98 which is around right here. Um, if you trace that, you will get to your gashes state. So that's your guys did. So you can either use your curve or use the boiling point and melting point given, but that's just another way of doing it. Okay, party last part here. Temperature where you have both liquid and gas. Of course. Um, you'll transition your transition from liquid to gas when you are boiling. Okay, so that would be where you're boiling. Point is, which is given right here in our problems or questions them. OK, so that temperature is 80 0.1 degrees Celsius.

So now we're going to work on problem 101 from chapter 13. This problem. We want to know about the freezing point of a nephew after cleaning solution A fencing, their events in solution of Napoli, uh, and were given some different parameters for benzene as well as the masses of each constituent. So first we need to calculate our moles of Salyut, which is moles of naphthalene. So we're told we have five grams of Napa lean. If we divide by the molar mass, we get the number of moles, which is 0.390 moles. So then we can calculate morality. It's equal to the moles of these Salyut divided by the kilograms of the solvent which is 0.444 and this gives us a value of 0.879 morality for Napoli. So we now know know the morality were given the freezing Thea Boiling point. Sorry the freezing point Depression constant so we can calculate DT. So the constant is 4.90 degrees Celsius per reality. The morality We just calculated a 0.879 and our Vantaa Factors one because NAFTA lean will not associate benzene and this gives us a value of 0.430 degrees Celsius. So we're given in the problem the typical freezing point of pure benzene, which is 5.5 degrees Celsius. So we need to go ahead and subtract our calculated value to get a value of five going 07 degrees Celsius as our new freezing point.

So we're looking at into molecular forces again where we have weekends molecular forces that exist in between solids as well as liquids. Where are gasses have broken away from these weak interactions. So we're looking for a melting point here. So we're looking for the relation between the liquid vapor pressure, which is p off fanzine at temperature T. This is as follows. So it fastly now 6.90565 subtract 1 to 11.33 over to 20 point 79 Several AT T. We see. So when we are triple point, all three exists of our phases will exist in equilibrium. So we have the following equation of about 6.90565 Take away 1 to 11 point, not 33 divided by 2 to 0.79 AD 30 degrees Celsius on this equals 9.846 Take away 2309 divided by 273 Ad temperature T which you two equations that we looked at earlier. Then we saw the temperature T, which is 5.3 degrees Celsius. And this is the melting point of fanzine and it is very close to the actual value that was given

So the relationship between the liquid vapor pressure p of benzene at temperature t is as follows. We have Log P and units of millimeters of Mark created 6.97565 Take away 1 to 11.733 divided by 2 to 0.790 t and units of Greece Celsius in the relation between the sublimation pressure of solid piece of the benzene at temperature T is as follows. Log P in units of millimeters of mercury is equal to 9.846 Take away 2309 Start by t in Calvin so we can rewrite the relation to in terms of the Celsius temperature log PT in millimeters of mercury is 9.846 Take away 2309 by the by T degrees C 273 So at the triple point, all three phases exist in an equilibrium within the system. So the vapor pressure of the liquid decorum benzene is equal to the vapor pressure of solid die chloral benzine. So this equates equation want to equation three. So we have 6.90565 take away 1 to 11.33 Try to buy 2 to 0.790 T degrees C is equal to 9.846 Segway to 309 divided by T degrees C at 273 so we can solve for temperature degrees Celsius. That is 5.3 degrees Celsius. Yeah, the melting point of benzene is 5.3 degrees Celsius, and this is very close to the actual value.


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