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1. (a) Microchip etching roughly follows the relation 𝑑 = 16.2 −16.2𝑒^−0.21𝑡 , 𝑡 < 100, where d is the depth of the...

Question

1. (a) Microchip etching roughly follows the relation 𝑑 = 16.2 −16.2𝑒^−0.21𝑡 , 𝑡 < 100, where d is the depth of the etch inmicrometers (µm) and t is the time of the etch in seconds. What arethe units associated with the numbers 16.2 and 0.21?2. A room of volume V contains air having equivalent molar massM (in g/mol). Calculate the mass of air that would leave the roomif the temperature of the room is raised from T1 to T2. Assume thatthe air pressure in the room is maintained

1. (a) Microchip etching roughly follows the relation 𝑑 = 16.2 − 16.2𝑒^−0.21𝑡 , 𝑡 < 100, where d is the depth of the etch in micrometers (µm) and t is the time of the etch in seconds. What are the units associated with the numbers 16.2 and 0.21? 2. A room of volume V contains air having equivalent molar mass M (in g/mol). Calculate the mass of air that would leave the room if the temperature of the room is raised from T1 to T2. Assume that the air pressure in the room is maintained at P0. Given • Ideal gas law PV = nRT , where n is the number of moles, P is pressure, T is temperature and R is the gas constant



Answers

A cylinder of volume 0.300 $\mathrm{m}^{3}$ contains 10.0 $\mathrm{mol}$ of neon gas at $20.0^{\circ}$ C. Assume neon behaves as an ideal gas. (a) What is the pressure of the gas? (b) Find the internal energy of the gas, (c) Suppose the gas expands at constant pressure to a volume of 1.000 $\mathrm{m}^{3} .$ How much work is done on the gas? (d) What is the temperature of the gas at the new volume? (e) Find the internal energy of the gas when its volume is 1.000 $\mathrm{m}^{3} .(1)$ Compute the change in the internal energy during the expansion. (g) Compute $\Delta U-W$ (h) Must thermal energy be transferred to the gas during the constant pressure expansion or be taken away? (i) Compute $Q$ , the thermal energy transfer. (j) What symbolic relationship between $Q,$ $\Delta U,$ and $W$ is suggested by the values obtained?

Okay. In this problem, we are looking at molecules rebounding off the walls of a container with length. Little well, We assume that we know the number of molecules in the container, and it's on. The gas in general is has a temperature of tea and a pressure p. We assume they're known for their first asked to show that the frequency of collisions with walls is expressed by this formula. We then in part B show that this frequency can also be expressed in this alternate form and lastly were asked to find actually estimated value for F for pressure at one atmosphere, temperature 20 degrees Celsius and air, which is soon to be a massive 29 times the proton mass. So let's start us off. So we start with the round trip time for a particular molecule. What's to say T round trip T. And so if we assume that the average time for a molecule to travel from one side of the box the other and back again is simply the average speed of the molecule, it's simply the average distance divided by the average speed could say something like that. Delta X, the total distance in the round trip divided by its you average speed of the molecules will say in the ex direction. And we can call this dealt X for our cube of length. L for a round trip is just two times the length of that cube divided by our average velocity, and we can use for a second fact the ideal gas law P V is equal to and Katie Ran is the total number of molecules kids, Bozeman, constant and everything pressure, volume, temperature. This So it must be the case that the number of molecules can be related to volume and pressure and temperature as so. And finally, you can replace the volume of this cube in terms of its length. PT times al cubed over Katie. This is useful to us because we can now say frequency of collisions will be simply the number of molecules times the reciprocal of the expected collision time, which is tr p t. R t. The total number of collisions is just gonna be the frequency of collisions per molecule, which is one of 30 times the number of particles. And we cannot replace, end with over other stuff and replaced you with it. This expression. So doing a little bit of multiple replacements. Weaken. Do this substitution. We could do this substitution. Doing some simplification. We're left with the expected result. So recall just a summarize. We used to get a frequency. We derived this round trip time for one particle hitting the walls. Expressed it in terms of the volumes, uh, side length of this cube volume and the average speed of a particle. And we also used the ideal gas law to express the total number of particles in terms of various other quantities. And then we use that to construct frequency of collisions. Okay, so that's part a Let's weaken box that now two for part B, you want to get you want to get the frequency in a different form. We can start with the fact that the they say the average velocity of a particle, he's gonna be the square root of its square. I mean, that's that seems self explanatory, not a big deal. We also know the average square this is from kinetic theory is equal to 1/3 of the average total average square. The average square in the extraction is just 1/3 of the total. That's an interesting result. We also know from kinetic theory that the average kinetic energy won't have and B squared average is equal to three halves. Katie, this is fundamental to kinetic theory. This result right here. So we now combined these three fax. You are a result from car day. Here's a result. From Friday we can replace the VX, the average VX. It's approximately this, but now we do a second replacement. You can express it in terms of a total average velocity, and then finally, we can replace it entirely with something involving the temperature of the gas, instead using this third relation, which would give us a route 1/3 times three Katy over m all that over, too. Always stuff on the outside, the square root remaining the same and we can simplify it down to get the expected results. He else weird all over roots for M k t. Okay, so we had to use three different relationships one and one someone trivial a second Not so obvious, but coming from kinetic theory and 1/3 as well. The fundamental the kidney Nike theory relating the average velocity, total velocity of particles to the temperature of the gas. OK, so that's our final relation. That's part B. Now, for part C, we often put some numbers. We have air. We're assuming that its mass is about 29. You were you is the proton mass sewing. The pressure is 1 80 m one atmosphere and we're sitting assuming the temperature is just 20 C or to 93 Kelvin So all we have to do it find the frequency is to simply input what we know into our expression that we have. We're gonna use the part be expression pl squared over route four AM Katie, start liking some numbers in 1.13 times 10 to the five pascal times three meters squared all over the square root of 4 29 Master proton 1.66 have send the native 27 kilograms Holtzman constant 1.38 in the name of 23 jewels for Calvin and our temperature and Kelvin to 93. So I need to extend that square it a little bit and that's when you do this computation. We're left with about 3.27 times 10 to the 28 hurts so extremely high frequency, and this is to be expected to have so many particles in the gas all hitting the wall relatively quickly. So it's not unexpected that we didn't have a huge number for our frequency and that's what it is, and that's it.

In this problem, given that radius is equal to one by two centimeter, which is equal to 0.5 centimeters. Changing it in meter. I can write the value like this and I can write the value of mass is equal to 15 A. M. U. So multiplying it by this value. So changing the value in kg. So on solving it to get evaluate 2.4915 multiplication and 10 to the power minus 26 Katie temperature here is given by the is equal to 21 C which is equal to 21 plus 2 +73 Calvin which is equal to 294 kelvin. And the value of P. F. Is equal to P. I by two. So pf by P I is equal to one by two which is equal to 0.5. On further simplification I can write the value of T. Is equal to long pierre by pr multiplication, be by ari Squire multiplication, understood. And by eight by K. B. The which is equal to long 0.5. Under the mode I forget to write the mode multiplication 1 to 5 m. Q by 0.5 multiplication. 10 to the power minus two meter. Holy square multiplication. Under the hood. 2.429 15 multiplication and 10 to the power minus 26 kg by eight by multiplication, 1.381 multiplication 10 to the power minus 23 multiplication, 294 Calvin. So on simplification I get the value of T is equal to 1712.92 2nd. It is because the there is a mode. That's why I forget the negative sign. Changing it in minute. I just multiplied by this. So final it is coming at 29 minutes approx. As 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.

All right In this problem, we were asked to look at the ideal gas law in the vendor roll equation of state for a scuba tank. So this is a four part problem. I've showed the 1st 3 parts, uh, in part they were just that were asked if 2300 leaders of air are brought into a tank. That's only 12 leaders large at a pressure of initially at a pressure of one atmosphere, and the temperature is 20 degree Celsius throughout the process. Show that the amount of moles in the tank is gonna be 96. Mole weaken. Do have the ideal gas law. Pretty simply, Part B then asked to use the ideal gas law where we know the number of moles in the tank and its temperature toe predict pressure in Part C were asked to use the venerable equation to predict a pressure in the tank where we assumed about some values. The parameters for air of aid is 0.1373 India's 3.7 to times 10 to the negative five, and finally, in park. Do you were asked to compute to confirm that the percent error is about 3% between these two. So I start us off for Friday. Brass used ideal gas law, which is PV equals, and Artie, Where are is the gas constant and number moles Piece pressure V is volume and his temperature. We know all these things have, except for end, which were asked to confirm so. And it would just be P V over R T Pressure in Pascal's is about one times 10 to the five we're giving a volume of which ends up being 2.3 meters. Tubed. Gascon sins always 8.314 and units are jewels. Permal, Calvin off and our temperature just 20 degrees is to 93. Calvin, remember, we have to convert to Calvin and Pascal's. Keep our units consistent. We're left with 95.64 mole, which is approximately 96 malls, which is what we were asked to confirm. That's it. That's not so bad for Barbie. We once again used ideal gas, Law said. Now we need to find pressure. So PV goes and Artie, we're given the number of moles in the tank in this case, and the temperature and the volume of the tank So pressure is just going to be put it over there. So pressure is just a party over V. So we go, we have to convert these volumes two meters cubed the point tank. And you just gave this 0.12 number. Moles, we assume, is 96. That's given in part B. Once again, the gas constant 8.314 and a temperature of 2 93 Calvin or 20 threes. All right. And we end up with about 1.9 times 10 to the seven Pascal Cool. Which is about 190 atmospheres. Remember, one atmosphere is about 10 to the five. Pascal. Cool. So you can see that this air in the tank is quite pressurize is almost 200 times the pressure. Uh, regular. All right, So I'm gonna go to another page for valuable equation, cause usually pretty large. Let's do that. All right, So remember the van rolls equation p. The predicted pressure for the mandibles equation is Artie over via Veran. My say it must be I always mess that up, and then it has a second term minus a over the over and squared Great and so we can expand this a little bit to make it a little bit easier to calculate things. This is equivalent to an arty over V minus N B minus and squared over V squared. Okay, so we're gonna use the exact same values as in part B for the volume. So just write that if volume is ah 0.12 meters cubes temperature is to 93 as before, Kevin and we have our values A and B from before. What was it? It was, uh, writer's point 1373 And I'm just gonna take the units, officer kind of messy and B is 3.72 times 10 to the negative five. We could put all these values in and compute p, which comes out to be about 1.9 times tender. That seven Pascal. All right, So far, so good. Cool. So, finally, our last part is, all we have to do is compare these two values. Um, and because they're so close to each other, actually, to use more precise values, get this percent air. I've just included the approximation toe like one or two significant figures, but we have to do a little bit better than that. I'll include the more precise values here. So we have a pressure from the ideal gas law as the pressure from the Vander Wal's equation or the pressure on the valuables equation. We're just getting a percent air. Here's were just comparing the pressure from Parsi and the pressure from party. All right, so now I'm gonna quote you some more precise values of these. We can actually get a good percent air. You have 1.9488 captain of the seven Pascal. That's ideal gas law. And 1.89 58 times 10 to 7. So I just listed these as 1.9 each in part B and C, and that's sufficient. But now we have to be a little bit more precise. Okay? And once we calculate this, we end up with 2.796% which is approximately 3% which is what we were asked to D'oh! And that's it. We've confirmed it. So the Van der Waals equation ideal gas law both give you pretty close estimates for pressure inside scuba tank. They're slightly off, but ah, pretty consistent with each other. So that's it. Just practicing using those two equations estate for the gas


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