what-is-the-maximum-vapor-density-possible-at-30circ-mathrmc-2

Name: what is the maximum vapor density possible at 30circ mathrmc 2
Uploaded: 2020-10-16
Duration: 1 min 34 s
Channel: Numerade
Description: What is the maximum vapor density possible at $30^{\circ} \mathrm{C} ?$

Question

What is the maximum vapor density possible at $30^{\circ} \mathrm{C} ?$

David Collins · Accepted Answer

question Number three asks you to determine the maximum vapor density that is possible at 30 degrees Celsius. It doesn&#x27;t tell you what units, and it doesn&#x27;t tell you how to determine this. It&#x27;s not a calculation. The key is understanding what you just read in this section. In this section you will find a graph that looks something like this. Where we have the water vapor density in units of moles per 1000 liters has a function of temperature in degrees Celsius. And if you look on this graph, you can see at 30 degrees we get a number density. That is about one point six somewhere around there, 1.6 to 1.7. So I&#x27;m going to say it&#x27;s about 1.6 six moles of water per 1000. Leaders of air

Averell Hause · Answer

So here the percent relative humidity percent relative humidity is defined as the vapor density divided by the saturation vapor density times 100%. And so we can say that, then the vapor density. I would be equal to the relative humidity percent multiplied by the saturation vapor density, and then this would be divided by 100. And so, for a relative humidity of 6% and a saturation vapor density of 51 grams 51.1 grams per cubic meter, we can say that the Baber density would be equaling again. 6% relative humidity 6.0 multiplied by saturation vapor density of 51.1 grams per cubic meter, divided by 100. And we have a vapor density of 3.0 Grant, 3.7 grams per cubic meter. This would be our vapor density. That is the end of the solution. Thank you.

Matthew Confer · Answer

the vapor pressure of water is 760 tour or one atmosphere at 100 degrees C. We know this because it&#x27;s the normal, a k one atmosphere boiling point of water.

Kevin Zaborsky · Answer

we want to determine what the vapor pressure of water is at a temperature of 100 degrees Celsius. We know that in order to measure the vapor pressure of liquid, we first injected into a tube containing mercury so that it will flow to the top of that tube, since it is less dense than mercury. So as that liquid evaporates water in this case, once it once it evaporates into the vapour phase, it exerts a downward pressure on that that Mercury column to change its height. And we can use this equation to help us determine what the vapour pressure is from that change in height of mercury. In the column, we know that on 100 degrees Celsius, this is equal to the boiling point of water. When we&#x27;re at the boiling point of a liquid, it means that the liquid and vapour phases are at equilibrium so that the rate of evaporation equals read a compensation. So therefore there will be no change in the height of mercury in the column. So that means that since we&#x27;re at the boiling point of water at 100 degrees Celsius, it&#x27;s vapor. Pressure is equal to the atmospheric pressure, and so the vapor pressure of water at 100 degrees Celsius is equal to the atmospheric pressure, which is one atmosphere.

Zachary Brauchler · Answer

This is problem 74 of Chapter 17. This problem wants to know how dense the electric energy can get in dry air before die. Electric breakdown occurs in the book on pages 6 27 and 6 28 They go over how to get the formula for electric energy density and that is you is equal to 1/2 die electric constant. Absolutely not. Times e squared. I go over to get this, but the book does it really well. So just go to those two pages to look at it. The electric field required to break down dry air is roughly three times 10 to the six volts per meter. In the dialectic, constant of air is about one plugging these values into this. We find that you is equal to 40 jewels per meter cubed. It&#x27;s in units of energy per volume because its energy density And that&#x27;s your answer. Good luck

Matt Ryan · Answer

problem, 24 has given us some di ethyl ether at 200 and 33 millimeters of mercury at 25 degrees Celsius. It&#x27;s just sitting in a flask and were asked, What&#x27;s the density of that gas? Well, this is a fun little twist on the problems that we&#x27;ve been working on so far, but we&#x27;ll be starting the same way with our good friend Piven ERT pressure volume, Mol&#x27;s gas constant and temperature and Kelvin out Density, as you may remember, is mass divided by volume. We&#x27;ve already got volume and technically we&#x27;ve got ah kind of a distorted version of mass. The moles I was thinking the same as the mass of our gas divided by its molar mass. This is the same thing as moles. Do you have grams, grams per mole? The grams cancel in the mold gets flipped up on top. So technically we have mass just like mass plus one. I brought a friend So all we need to do is rewrite our Pinar are massive, our gas divided by Moeller Mats rt and now we&#x27;re ready to go. We want to make sure that we bring this volume to the this side here, so we&#x27;ll multiply this side by one Overbey aside by one Overbey and we want to maneuver. These guys are more mass, our gas constant in our temperature in killed in to the other side. So we multiply both sides here Those cancel we&#x27;ll multiply both sides by the reciprocal of this so molar mass on top you Oh, I&#x27;m I&#x27;m thaw we for that. That was glows and our gas constant in temperature and Calvin on bottom. And because we&#x27;re good mathematicians, whatever we do to one side we do to the other, I am. Go away, you&#x27;re out of here. And our final equation that will be plugging our numbers into is gonna look something like this. Our mass divided by volume massive RGs, which is density is gonna be equal to our molar mass of our diable either times the pressure it&#x27;s currently sitting at, we&#x27;ll have to do a unit conversion on that to get into atmospheres divided by the gas, constant times, the temperature and Calvin. So we don&#x27;t know the volume. Ah, but that&#x27;s OK because we&#x27;re being asked to solve for density in this case. Um, but we have everything else that we need. Let&#x27;s go ahead and plug those in the molar mass of diet awful either is 45 0.0 for grams promote, and we will need to convert ah this into a T M&#x27;s as almost every gas constant value you&#x27;ll find. It has its pressure portion in ATMs, so we&#x27;ll just use our conversion factor for every millimeter of mercury Earth. One atmosphere is 760 millimeters of mercury. Ah, this will convert. Let&#x27;s see for a team that&#x27;s our conversion factor so we can see that are millimeters of mercury will cancel and this will get flipped up on top and we&#x27;ll have our units and atmospheres. All of this gets divided by our gas constant. I am using the leaders version 0.8 to 05 leaders times a. T. M. You Yeah, yeah, Calvin Mole. And this is multiplied by our temperature, which we will need to convert into Kelvin&#x27;s. So to 73 plus 25 plug all this into our calculator and we find what the gases mass divided by the volume or its density is equal to. In this case, that will be 0.5 six for seven grams for leader. Looking back at our Sig Figs, we have to. So around this 2.56 grams per liter, this job.

Problem

Is it possible for the water vapor density in the…

Problem 3 Easy Difficulty

What is the maximum vapor density possible at $30^{\circ} \mathrm{C} ?$

Answer

about 1.7 mol per 1000 $\mathrm{L}$

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about 1.7 mol per 1000 $\mathrm{L}$

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Angelica M. StacyLiving by Chemistry

Theodore D.

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Morgan S.

Jake R.

Angelica M. Stacy

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