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Q 2a. (10 Points) Use the van der Waals equation to calculate the pressure of nitrogen gas at 273.15 K and molar volume of 22.414 L mol Compare with the pressure of...

Question

Q 2a. (10 Points) Use the van der Waals equation to calculate the pressure of nitrogen gas at 273.15 K and molar volume of 22.414 L mol Compare with the pressure of an ideal gas at the same temperature and molar volume. 1.37 barL2 mol- b - 0.0387 Lmol-1Q2b. (10 Points) The coefficient of thermal expansion of ethanol equals 1.12 * 10-3 K-1 at 20 %C and 1.000 atm_ The density at 20 %C is equal to 0.7893 g.cm Find the volume of 1.000 mol of ethanol at 10.00 OC and 1.000 atm_ The molecular weight of

Q 2a. (10 Points) Use the van der Waals equation to calculate the pressure of nitrogen gas at 273.15 K and molar volume of 22.414 L mol Compare with the pressure of an ideal gas at the same temperature and molar volume. 1.37 barL2 mol- b - 0.0387 Lmol-1 Q2b. (10 Points) The coefficient of thermal expansion of ethanol equals 1.12 * 10-3 K-1 at 20 %C and 1.000 atm_ The density at 20 %C is equal to 0.7893 g.cm Find the volume of 1.000 mol of ethanol at 10.00 OC and 1.000 atm_ The molecular weight of ethanol is 46.07 &mol-1.



Answers

At 273 $\mathrm{K}$ and $1.00 \times 10^{-2}$ atm, the density of a gas is $1.24 \times$ $10^{-5} \mathrm{g} / \mathrm{cm}^{3} .(\mathrm{a})$ Find $v_{\mathrm{rms}}$ for the gas molecules. (b) Find the molar
mass of the gas and (c) identify the gas. See Table $19-1 .$

In this example, we're going to take a look at a model of a non ideal gas known as the Vanda balls model, which is an experimentally based model. Um and we're going to compare it to the ideal gas, which assumes that the gas particles Have zero volume or or points and that they do not interact other than somehow kind of bouncing off of each other. So pressure times the volume of the guess is the number of moles in the gas times, the gas constant times the temperature is true for the ideal gas in the Van der balls model. The left hand side of the equation is um yeah, uh, changed a little bit. So the pressure is raised. Um bye. A small amount supposedly and the volume is lowered by a small amount supposedly. And that right hand side has not changed any. So you'll notice in this model there are two parameters, there's an A and B parameter and is still the number of moles and the V is still the volume and he is still the pressure. But what are these parameters? Well, they have a physical meaning. The B is probably a little bit easier to understand. It comes from the size of the particles, the actual physical size. Over on the left here, I have some data on some real gases. So there are three noble gases for comparison and then three gases that are molecules occurring in our atmosphere. And basically you see the trend that the B parameter that's used in the Van der Waals model increases as the number of um nucleons in the nucleus of the particle or the particles atoms if it's a molecule uh that as that mass number increases, so does the B parameter. Now mass is not the same as physical diameter, but it is true that the more particles you have in the nucleus of your atoms um that the electron cloud around the nucleus grows in size as well. Um Now the other parameter, the a parameter is related to the inter atomic attractions um which are typically considered to be weak uh in a gas, even though there there and how you see those inter atomic forces operating is with the boiling point. So I've written down the boiling points of these various gases, the temperature at one atmosphere basically at which they transition from the liquid to the gaseous face. And you can clearly see that the pattern is that the larger A. Is uh the higher the boiling points are meaning that the inter atomic forces harder to break apart to turn the substance into a gas. So for our example here, we're going to take nitrogen and two. And I'm going to go ahead and write down it's A and it's B. And the proper unit for the A. C. System. So it's a parameter Is a .14. If we convert the parameters above into pascal's for pressure, m3 for volume and we still have moles of gas, which is kind of unit list, but not really. We have to remember that a mole is a large number of particles and we have a B parameter that again converts nicely into the metric units. Yes, I standard. Um and that would be in cubic meters per mole and we're going to take our gas. It will be at 10 atmospheres of pressure And I'll go ahead and write that 10 atmospheres in terms of past gals. So that would be one oh 13 Times 10 to the 6th Pascal's So 10 atmospheres. Um the volume, we're going to Uh start off with leaders 2.0 leaders. It's easier to imagine that as a volume, but we do want to convert it to cubic meters and that's easy enough to do Factor of 1000. So to Times 10 to the -3 m3 and we're going to try to compare the temperature of the ideal gas to what the vander balls prediction would be for these same conditions. Um And let's see, I'm not going to put all the calculations in there, but the left hand side on the ideal gas model Gives us 2.026 1000 jules and 8.31 jules per kelvin on the other side times temperature. So we can definitely solve that for temperature. Um and we get about 244 Kelvin roughly to 43.8 kelvin. And notice that this is a fairly high temperature. It is well, maybe not a room that you would like to be in um but you know, certainly within the realm of human survivability, but this is much higher than the boiling point of nitrogen to so we would expect that nitrogen um even with the Van evolves model would be acting fairly much like an ideal gas. But let's see for the same conditions, we see that the pressure term raises the temperature and the volume term lowers the temperature and let's just see what happens um With this. So again, I won't bore you with the details, But the pressure gets raised 1.4, 8 Times 10 to the six pascal's and the volume gets lowered um to 1.96, 1 Times 10 to the -3 cubic meters. And that's still equal to 8.31 jules per Calvin times the temperature. And we can solve all that for the temperature, but we can see definitely the competing effects that are going on and that winds up giving us a temperature close uh huh. To what we expect from the ideal gas law, it's a little bit higher. So the pressure term does win out, meaning that the inter atomic forces are more important in the model at this set of conditions, then the fact that these are particles with extended sizes.

And this question. We're asked to calculates the pressure using the ideal gas equation and the Vander Waals equation of 1.5 moles of sulfur dioxide at 298 cope in at varying volumes. So I already have the ideal gas law and on the left and the Vendor walls equation written on the right, let's go ahead and begin first. Using the ideal gas law, we have 1.5 times 0.8 to 1 times 298 Calvin over 100 leaders, and this comes out to be 0.367 atmospheres of pressure. I'm gonna go ahead and copy and paste the Vander Waals equation. So we have pressure equals 1.5 moles. Time. 0.8314 times 298 Calvin divided by 100 leaders minus 1.5 times 0.0 879 minus 1.5 squared time. 7.857 over 100 squared does go ahead and plug this into our calculator. So this first part is coming. Come out to be 0.372 and the second part is going to come out to the 0.18 So the pressure based on the Vander Waals equation is going to be 0.3702 Okay, so now let's go ahead and continue. The only thing which ended here is going to be the variables. So I'm gonna copy and paste this. And instead of 100 leaders, we now have 50. I'm gonna go ahead and plug this into a calculator and we get 0.734 atmospheres. And remember, when we using the Vander Waals equation are, pressure comes out and units of bars. Let's go ahead and do the same thing for this site again. Instead of 100 leaders, we have 50. So a pressure I was going to be 0.745 minus 0.71 which comes out to be 0.73 79 bars. Next we have 20 leaders. So we have 1.835 atmospheres on this site, 1.871 minus 0.0 for four. So our pressure is going to be 1.8 27 bars. Lastly, we have 10 liters. So our pressure here is going to be 3.67 atmospheres, 3.60 bars. So now we're asked another question. We're also asked under which of these conditions is the pressure calculated with the ideal gas equation within a few percent of the one calculated with the Vander Waals equation. And if we take a look, we actually see that as we go higher in our volume, we actually get closer the pressure between the two equations. So let me write this down. As the volume increases, the pressure gets closer.

Okay. So this question tests your understanding on the boiling point and other some parameters relates to the space transition. So it says a tank of gas 21°C.. And as the pressure of the one A. T. M. Using the data in the table answered the following question, explain your answer. Yeah. So a question a is if the 10 contains carbon tetrachloride is a liquid state. Also President. So first, if you want to look at whether there's a liquid state, the first thing is to look at the boiling point. It says the boiling point at 1 80 M of the cf four is yeah minus 128° C. Which is far below. Okay. The 21° C. Which means most case it will be gases because the temperature, the environmental temperature is much higher than the boiling point. So in most cases it would be bored. And another way to kind of like testify that is to look at the critical temperature. The critical temperature tells you as this temperature. When the temperature go beyond this temperature, there's no way to just simply um mhm. Condense the gas phase into liquid to simply adding prep by simply adding pressure. So says the critical temperature will say F four is -46°C.. As far below 21. The 21°C.. is much higher than that, which means that 21°C.. Even you are able to change the pressure Still, the CF four will not be liquefied. We're not condensed liquefied. So, I would say for the question a there would be no liquid state presence and for the question be ask you for the detain, say four H 10. So generally you say the bowling point is actually still Lower than the environmental temperature 21°C.. So that means that it's probably uh in the gas phase. However, if you look at a critical temperature, You will find that it's much higher than the 21, which means it's possible if you want, you can try to apply or increase the pressure of the gas tank, so that actually pertain can be liquefied. So it is possible that will be liquid state uh presence in the face if you have a higher pressure than a. T. M. And this is the answer.

In this problem working with wonder pills question upstate, I would respond there. It will be expected to be. The assumption is that we're witnessing that and are plenty. So the right inside is not a function off Pete. So let's take their little both sides. Expect Pete and we have one. Waas ddb oh, in script A You buy. We screwed times three months and we were using for a cool plus p plus inscribed a wee spurred times D B E T is equal to zero. So doing just appreciation. We have one minus two in spirit times A tense meeting. Leggett third TVP Hey times we minus and B was P plus end script. A light we skirt Tench DVD is zero, so let's grew older. Term to DVD. Feel one side in this hole terms without DVD grown other side. Let's write this one is GDP times to inscribe a digital negative hard times, then B minus B plus He waas been scraped times any time to physical and be my speed. So from this we see that on TV. PeopIe physical N B minus B Derided by to end Scrag times eight times 32 negative. Third and B minus three plus p wasp Insert a beater donated to All right Now we can multiply both sides for numerator and even Norman, and you would reach in the neck with a positive to get to go, mister and write the answer as DVD is able to end. B minus three. Time Beat You divided by, too. Then Jew babying plus Pete and beat you minus and spurred. A. This is the answer to First Part, a four part fever given bellies for N B A, um, envy and rest. If I dd forgiven Melodies you b plus everything. NBC that the answer is or DVD P for the given conditions is named for 0.44 leaders, for that was very pressure.


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