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3.) When Grandma seals the lid on her pressure cooker theinitial temperature is 272 K and the pressure is 0.850 atm. Aftershe heats it to 393K, what is the pressure...

Question

3.) When Grandma seals the lid on her pressure cooker theinitial temperature is 272 K and the pressure is 0.850 atm. Aftershe heats it to 393K, what is the pressure in the cooker?Group of answer choicesa. 1.23 atmb. 0.950 atmc. 1.37 atmd. 0.658 atme. 2.57 atm4.) A latex rubber weather balloon is filled with 75.0ft3 of helium at sea level where the pressure is760.0 torr and the temperature is 298 K. The balloon is releasedand rises to an altitude of 16,500 feet. At this altitude,the pressure is

3.) When Grandma seals the lid on her pressure cooker the initial temperature is 272 K and the pressure is 0.850 atm. After she heats it to 393K, what is the pressure in the cooker? Group of answer choices a. 1.23 atm b. 0.950 atm c. 1.37 atm d. 0.658 atm e. 2.57 atm 4.) A latex rubber weather balloon is filled with 75.0 ft3 of helium at sea level where the pressure is 760.0 torr and the temperature is 298 K. The balloon is released and rises to an altitude of 16,500 feet. At this altitude, the pressure is 403 torr and the temperature is 249 K. Assuming no helium is lost, what is the volume of the balloon in ft3 at 16,500 ft? Group of answer choices a. 36.8 ft3 b. 25.7 ft3 c. 58.0 ft3 d. 118 ft3 e. 131 ft3



Answers

A $10 \mathrm{dm}^{3}$ SCUBA cylinder is filled with air to a pressure of $\left.300 \text { atm at a temperature of } 20^{\circ} \mathrm{C}(293 \mathrm{K}) . \text { (Section } 8.6\right)$ (a) Calculate the amount in moles of gas in the cylinder, assuming the air behaves as an ideal gas. (b) When the diver jumps into cold water at $278 \mathrm{K}$, the pressure gauge shows an alarming drop in pressure. Explain the reason why and calculate the new pressure inside the cylinder. (c) In fact, the compressed gases do not behave as ideal gases. Explain why. Use the van der Waals equation (for air, $a=0.137 \mathrm{Pam}^{6} \mathrm{mol}^{-2}$ and $b=3.7 \times 10^{-5} \mathrm{m}^{3} \mathrm{mol}^{-1}$ ) to show that the amount of air in the cylinder is 115 mol. In view of your answer to part (a) above, what are the implications of this for divers?

A weather balloon is being launched from sea level, where the atmospheric pressure is equal to one atmosphere and the temperature when converted into Kelvin is equal to 293. The volume of the balloon wanted a shot off initially is equal to 200 leaders. We want to determine in parts A and B what the final volume of this balloon will be once it reaches two different altitudes. In part C of this problem, we want to know whether the balloon will rupture at either one of these altitudes. We begin by looking at the ideal gas law and we know that the pressure, the volume and the temperature are all changing with time as a balloon, a sentence from initially the surface of the water level up to 20,000 meters in 50,000 meters above sea level, so n and are the number of moles of air in that balloon as well as the ideal gas constant. We're going to be the fixed values in this problem by the pressure, volume and temperature change. And when that's the case and we divide over the temperature to get all the constants on one side, we see that the product of the pressure and the volume divided by the temperature of the gas is equal to a constant value. And when that's the case, we have the combined gas law where u one V one over t one equals P two be to over t two. We have information in each part of the problem to solve her v two since we know what the values of the other variables are so we can rearrange the the combined gas law, the salt for V two. Using this equation, he one v one t two divided by he to t one. So starting with part A, we are told what the pressure and temperature are of the balloon. When it is 20,000 meters above sea level, we use the comm combined gas law and the values given in the problem to solve for that final volume. So again, the initial pressure at sea level is one atmosphere. In that initial volume of the balloon was 200 leaders the final temperature given in part a this 220 degrees kelvin, and this is divided by final pressure, which, when we converted the millimeters of mercury and toe atmosphere came out to final pressure of 0.83 atmospheres and again the initial temperature is 293. Kelvin. We should have to sig Figs in our final answer and when we calculate this out, the volume of the balloon when it is 20,000 meters above sea level should be equal to 1.8 times 10 to the third leaders. And that's the answer to part a. Part B is very similar to part A, except we have. The balloon is now ascended to 50,000 meters above sea level, so we're given to different final pressures and temperatures. But the algorithm algorithm that we used to solve for the final volume is the same as in part a where again, the initial pressure is one atmosphere. The initial volume of the Belounis 200 leaders still the final temperature and part B is 270 killed in final pressure when converted from millimeters of mercury in tow. Atmospheres is zero point 00105 atmospheres and we want to play that by the initial temperature, which again was 293 Kelvin again. We should have to Sig Figs in our final answer and when we perform that calculation should get 1.8 times 10 to the fifth leaders as a final volume for the balloon when it is 50,000 meters above sea level. Reports see, we're told that the balloon will rupture if it's volume exceeds 400 leaders and we want to know if and either part A or B, the balloon would burst. What we see in either case that V two and Part A is greater than 400 leaders and V two in Part B is also greater than 400 leaders. So the answer to part C would therefore be yes. In both cases, when it's 20,000 meters above sea level on 50,000 meters above sea level in parts A and B respectively, the balloon will burst because for both of those, the final volume is greater than 400 leaders

This is problem for from Chapter 11 section for and we have this balloon that is filled with 240 million years and the temperature is 25 degrees Celsius to start with and the pressure is on one atmos. Good pressure. Then we carry this boon of a mountain when the pressure and the pressure drops 2.75 atmosphere pressure, but the temperature still remains at 25 degrees Celsius, so a is asking what changes. So if temperature is staying the same, we can bet that this is a pressure and volume relationship. So pressure and volume or going to change and get temperature in the number of Mali Krul's are going to stay the same. So be is asking if if air pressure is going to decrease now. Air pressure did decrease because we saw that when you went up the mountain, it dropped from one 2.75 See is asking what happens to the value, so see we'll have an increase in volume since we have a decrease in pressure to the volume in the balloon will expand due to less pressure on the outside. So he is a calculation and we're trying to find the new volume. So we're gonna use this equation. P one times of the one is equal to okay to times. Me too. So we're trying to find me too. So we have p one of 1.0, a t m times the volume one, which is 200 and 40. You know, as it's gonna be equal to the new P two, which is 20.75 and we don't know the to is someone put X here. So when you multiply p one of the one together, you still get to 40 and still be equal 2.75 a t 10. So when your song for eggs divide to 40 by 0.752 X should be equal to 300 and 20. And so this is the new volume when going up the mountain

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

And this problem. We're trying to figure out how much mass a hot air balloon can lift. Um, so the total mass that the balloon left is the difference between the mass of air displaced by the balloon and the mass of the gas and five, the balloon. Um, we're being told to consider a hot air balloon that has a ah spherical shape, um, in a diameter of five meters, where the air is heated to 65 degrees Celsius in the surrounding air temperatures, 21 degrees Celsius and the pressure both within and outside of the balloon is 745. Tour. Um, first thing we need to do is figure out the volume off our hot air balloon, and the volume of the spear is 4/3 pi are cubed on, and since the diameter was five meters than the, um, radius is going to be 2.5 meters. And so when we substitute in the 2.5, we get that our volume is 65 0.45 meters cubed, which equals 65,000 450 leaders of air within the hot air balloon. We're also gonna want to go ahead and convert our temperatures into Calvin. So of its 65 Calvin started 65 Celsius 65 plus 2 73 which gives us a temperature of 338 Kelvin and then for the 21 degrees would do the same thing. 21 plus 2 73 is 294 Kelvin. And then finally, we're gonna want to convert our pressure from toured the atmosphere. So we divide the 7 54 by 7 60 and we get that the pressure is going to be 0.98 a t m. Okay, Um, so the first thing we're trying to figure out what mass? The Blinken lift, assuming that the average Mueller mass of air is 29 grams per bowl. So first we're gonna figure out how many moles of air we have inside of our balloon. So to do that, we're gonna use PV equals at R T and solving it for M so n equals 0.98 which is the pressure times 65,450 which is the volume divided by our gas constant. 0.8 206 times are temperature and the temperature inside the balloon is 338 Kelvin. So we get that We have 2000 300 and 12.5 moles of air inside of the balloon. Now, since we have, we were told that the mass of the air, the mill a massive airs 29.0 grams per mole. We can now figure out how many grams the balloon is able, or how many grams of air is it inside of the balloon. So Well, get on Muller Mass is going to be the 20 2312.5 moles times 29 which gives us 67,000 63.39 grams. Okay, so now that we know the mass of the air inside of the balloon, we now need to get the mass outside of the balloon. So we're gonna do the same thing we're gonna get from the number of moles, which is gonna be your 0.98 because they both have the same pressure. We're gonna use the same volume because the the air that's getting pushed out is the volume of the balloon times are gas constant eight 206 times are the temperature outside and the temperature outside of the balloon is our 2 94 So, times to 94 and we get that we have a total of 2658.6 two moles and then to get our mass were again gonna multiply that number 658.62 times 29 and we get a total of 77 those in 100 grams. So to find the mass that the balloon can lift, it's the difference between those two numbers. So we're gonna take our roughly seven point 71 times 10 to the fourth grams minus are 6.7 zero times 10 to the fourth grams, and we'll get 1.101 times 10 to the fourth grams can be lifted by the balloon. Now, for part B, it's Afghanistan to figure how much the balloon could lift if it was actually filled with helium at 21 degrees Celsius in 700 and 45 tour. Um, assuming the same volume, etcetera. So let's go ahead and just clear up all this stuff, okay? and start fresh, so their volume is gonna be the same. Um, the mass of the air is going to be the same, but now we need the molar mass of helium. So molar mass of helium is going to be from the period table 4.3 grams per mole. So we're gonna do the same thing we we had from our previous part one apart. A. We know that the the mass that of air on the outside of the balloon is seven point 71 times 10 to the fourth, uh, grams. So now we can go ahead. And if we find the number of moles for helium in the situation, Megan, then just subtract that from the mass of the air. So the number of moles of helium is gonna be our 0.98 because it says the same pressure. We have the same volume, the 65 4 50 and we're gonna divide that by our gas constant and our temperature, which is still is gonna be the same as the temperature of the air. So to 94. And we get that that is, um, acquits to 2000 658 0.62 Moles of helium. And we're gonna take that number 6 58.6 to multiply it by our 4.3 which is the Moler Massive helium. And we get 1.6 times 10 to the fourth grams. So if we subtract that from our 7.71 times 10 to the fourth minus 1.6 times 10 to the fourth, we get an answer of 6.65 times 10 to the fourth grams can be lifted by the balloon if it was filled with helium. And for the final part of this problem, we're being asked to consider the same situation of group one, Um, except her from part A. Except this time we're taking into consideration it being in Colorado, um, with a pressure of 630 tour. So we're gonna have to go through that same process we started with, but this time changing our pressure from 7 50 Our new pressure is 630 tour. So divide that by 7 60 and you get that the new pressure is 0.83 a. T. M. so going through our problem again looking for the number of moles within the balloon we have 0.83 times our volume. 4 65,050 divided by our gas. Constant times are temperature. We're gonna use the same temperatures as we did in part a. So the number of moles inside the balloon equals 1.96 times 10 to the third. Moles take that times 29 and we get if our a mass of 5.68 times 10 to the fourth grams time and then doing this again for the air because again, the number of most for the air is also going to change. With the change in pressure 0.83 times 65,000, 450 divided by 0.8 to 06 times are temperature of to 94. We have that. We get 2.25 times 10 to the third malls. That's why that 2.25 times 10 to the third times 29 and we get 6.52 times 10 to the fourth, um, Grams and then finally tracked those two numbers from each other. So 6.52 times 10 to the fourth, minus five point 68 times 10 to the fourth. And we get a massive 8.3. Whoops. Sorry. 8.4 times 10 to the third grams.


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