imagine-you-fill-a-balloon-with-air-to-a-volume-of-240-mathrmml-initially-the-air-temperature-is-2-2

Question

Imagine you fill a balloon with air to a volume of 240 $\mathrm{mL}$ . Initially, the air temperature is $25^{\circ} \mathrm{C}$ and the air pressure is 1.0 atm. You carry the balloon with you up a mountain where the air pressure is 0.75 atm and the temperature is $25^{\circ} \mathrm{C}$ .
a. When the balloon is carried up the mountain, what changes? What stays the same?
b. The air pressure on the outside of the balloon has decreased. Can the air pressure on the inside decrease so that the pressures are equal? Why or why not?
c. What happens to the volume occupied by the air inside the balloon? Explain your thinking.
d. Solve for the new volume of the balloon.

Marissa Turner · Accepted Answer

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&#x27;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&#x27;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&#x27;re trying to find the new volume. So we&#x27;re gonna use this equation. P one times of the one is equal to okay to times. Me too. So we&#x27;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&#x27;s gonna be equal to the new P two, which is 20.75 and we don&#x27;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

A. Elizabeth Hildreth · Answer

problem. 1 65 says you have a helium balloon at one atmosphere and 25 degrees Celsius. You want to make a hot air balloon with the same volume in the same lift of the helium balloon. So assuming the air is 79% nitrogen and 21% oxygen, would the temperature in the hot air balloon have to be higher or lower than 25 degrees Celsius? So, um, in order to ah, think about this. The molecular weight of helium is about four grams per mole, while the molecular weight of air is about 29 grams for mole. Because air, um, is is heavier than helium, the temperature would have to be higher than 25 degrees Celsius to be less dense than the surrounding air, Um, or unequal density to helium at basically standard temperature and pressure. Well, I guess not standard pressure, but not temperature. Um, so the air in the balloon in this case will have to be heated to achieve the same density and thus achieve the same lift. So part B of this question then says to calculate the temperature of the air required for this hot air. Belene too, um, provide the same lift its helium balloon. And so again, the pressure given by the problem is one atmosphere and temperature is 25 degrees Celsius. So we can start this problemas many of our problems in this chapter with E ideal gas law equation where we&#x27;re solving for a number of moles off gas required, um, to fill these balloons. So the pressure and this problem this one atmosphere and we weren&#x27;t given a volume for this balloon. But just to make the path easy, I&#x27;m just going to go ahead and assume that the volume is one leader so you can plug in the ideal gas constant and the temperature We were Kevin, just 298 Calvin. And this gives us 0.0 for 08 moles of helium in the balloon. So because lift in, this problem is defined as the difference between the massive air displaced by the balloon and the mass of gas inside the balloon. What we need to do here is figure out the mass of helium inside of healing bloom. So, as I said before, we know that the Miller mass of helium is about four grams per movie, which tells us that the helium balloon has approximately 1.163 grams, um, of gasoline inside. So now we can use this information, Thio, um, to figure out, um, how much air we would need, um, and sign of our hot air balloon. And you sat to figure out the temperature. So as I said before, um, the Mueller massive air, it&#x27;s about 29 grams per mole, miraculously, 28.84 grams per mole. And so if we have, um, the requirement of 0.163 grams inside the balloon and we know that air is about 28.84 grams for mold, we&#x27;ll need 0.565 moles of air to match the helium balloon so we can plug this information then into the ideal gas equation, solving for temperature. Since our goal is to figure out the temperature of the air required to match the lift of this helium balloon. And again we could just pluck in the information that we have from a problem on the number of moles of air required Thio inflate this balloon. So our pressure is one atmosphere again we want the same volume. So we&#x27;ll just do one leader. We can plug in, um, number of moles here, this 0.565 and then we can plug in our ideal gas constant. So what that gives us is 2156 Calvin or a taking into consideration the proper number of significant figures 2.15 times 10 to the third Calvin.

Kevin Zaborsky · Answer

Hey, everyone. So today we&#x27;re going to look at this problem that asks us about Boyle&#x27;s law, and I&#x27;m going to begin by drawing a diagram of the problem statement so that we can get a better sense of what&#x27;s going on and what the problem is asking us to find. So we have this balloon that we filled with air. We&#x27;re gonna take it down with us as we scuba dive under the water to a final depth of 30 feet. Well defined this initial state before we take the balloon down with us under the water is number one in the final state, when we achieve a depth of 30 feet under the water as state number two. Now the problem asks us in part A. What would happen to the volume of the balloon? Meaning would it get bigger or smaller or stay the same size as we move from that initial to final state bringing the balloon down with us underwater. In part B of this problem asks what would happen to the pressure inside the balloon as we make that seem changed from the initial to the final state. So whenever I&#x27;m solving a gas log problem. I always find it helpful to start with the ideal gas law, which is PV equals NRT now T is the temperature of the gas and we&#x27;re told from the problem statement that this is constant. Are is the ideal gas constant and therefore by definition, is also a constant. Now, if we look at our visual again of the balloon and we know that end is a number of moles of air inside the balloon, the moles can be related to the number of molecules of air inside the balloon you&#x27;re of a gaja is number. Therefore, we can arbitrarily assume that there are five air particles inside this balloon and therefore the number of moles, and does not change either, because the race side of this equation is constant. The product of the pressure and the volume of the air particles inside the balloon also have to be equal to that same constant. Therefore, any increase in pressure must lead to a decrease in the volume to maintain the same constant value when the two quantities are multiplied together. So this inverse relationship of increasing the pressure leading to a decrease in volume and vice versa is known as Boyle&#x27;s Law and another way to present this is P one V one equals p two v two. We&#x27;re again one into in this scenario, refers to this initial and final states of taking the balloon down underwater for the second state and before that for the 1st 1 We know that from the problem statement, the final pressure is two atmospheres and we can assume that the initial pressure is equal to the atmospheric pressure, which is approximately one atmosphere. We therefore know that the pressure increases as we move from 1 to 2. This also makes sense because as we bring the balloon further down under the water with us, the force that the water from above exerts on the balloon will continue to increase as we get to a lower and lower depth under the water. Therefore, we know that the pressure increases as we move from 1 to 2 and from this relationship to find by Boyle&#x27;s law, we can therefore say that the volume of the balloon decreases. It gets smaller and that is answer to part a report. Be we again. Look at that same visual of the balloon with the five air particles in it before we bring it down underwater. And we know from part A that the final state of this balloon will have a smaller volume in the initial state. And pressure can be defined as a force that these air particles inside the balloon exert on the inner walls of the balloon. And because in the second state there is a smaller volume for those air particles to occupy, the force that they exert on those on the inner inner walls of the balloon will be greater due to the smaller volume, and therefore we can conclude in part B. The pressure inside the balloon will increase since again, the volume in the second state is smaller than it was in the first state.

Kevin Zaborsky · Answer

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&#x27;re going to be the fixed values in this problem by the pressure, volume and temperature change. And when that&#x27;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&#x27;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&#x27;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&#x27;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&#x27;re told that the balloon will rupture if it&#x27;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&#x27;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

Ronald Prasad · Answer

in this common classroom demonstration of cooling a balloon with liquid nitrogen lists. First, calculate the actual are Sorry. Let&#x27;s calculate the expected volume of the balloon upon cooling toe 196. If you&#x27;re Celsius, so will first have to calculate the moles of gas present in the balloon. Using our ideal gas formula. P V over R T pressure is 0.998 atmospheres. The volume initially is 2.95 leaders in the balloon R 08 to 06 leaders atmospheres per mole Calvin and the balloon is initially at room temperature 25 degrees Celsius. This would yield 1 to 0 moles of gas present in the balloon. Now we&#x27;re going to calculate the, um volume of gas expected upon cooling. We could do this by again using our ideal gas formula and r T overpay Number of moles is 0.1 to 0 moles 0.8 to 06 leaders atmospheres per mole. Kelvin Negative 1 96 degrees Celsius is 77 Calvin over a pressure of 0.998 atmospheres, and the expected volume works out to be 0.76 leaders. The next part of the question. We&#x27;re told that the volume of the gas that is actually observed upon completing the demonstration is actually 0.61 leaders, so we see that our volume expected is higher, So the value three observed value is less eso the observed value. Our volume is less than the expected volume because at lower temperatures, the collisions between the molecules occur with less kinetic energy and weak attractions do affect the collisions with the surfaces of the container, thereby lowering the pressure compared to the expected value.

Marissa Turner · Answer

This is Question five from Chapter 11 section for and it&#x27;s question on the relationship between pressure and temperature. So we have a glass bottle that has 180 no leaders, and after start there&#x27;s 25 degrees Celsius and the air pressure is one atmospheric pressure. And then after that, we take the class bottle up to a mountain where the pressure is 0.75 atmospheric pressure and the temperature drops to five percent. So in a is asking us what happens to the volume. And so in a the air pressure outside the glass bottle decreases, but the volume remains unchanged since nothing from the problem actually change from it. When you read it from above, you only saw that pressure and temperature change. So what isn&#x27;t it? Volume stays. Volume stays the same. Actually, I&#x27;m changed. Okay, for be. It&#x27;s asking about the temperature, and that&#x27;s the temperature expected to be court down to 5% of us up. And that is a yes because, actually, in the problem, it says that one going up the mountain, it drops from 25 degrees Celsius to five degrees Celsius, so see it asking what happens to pressure. So if temperature decreases, pressure is also going to decrease because temperature and pressure has a direct relationship. Okay, and d we&#x27;re trying to solve for the new pressure. So we&#x27;re gonna use an equation which is one of use key. One over team won is equal to p two and she to. But before we can use it, we actually have to convert the temperatures to Kelvin, cause right now it&#x27;s in Celsius. So the kelvin is 200 approximately 273. She is gonna add that on to the Celsius. So we have to 70. We have 25 degrees Celsius in the ads. Who&#x27;s 73? I was gonna give you to 98 and the five is going to give you to 78. Someone equal this. The pressure at the store is 1.0 80 And the pressure, the temperature of the store is going to be too 98 kelvin. And this is gonna be equal to we don&#x27;t know the pressure. The 2nd 1st of them. For next year, it&#x27;s gonna be equal to X over to 78. We can actually cross multiply for this to find it. So one times 2 78 it&#x27;s still to 78 Senate of it, too. 78. And we have to 90. So we can actually divide this into 78. Over to 98. Go to the final question is going to be This is gonna come out to be a point 93 80 m. So this is the pressure when going up the mountain.

Problem

The air inside a 180 $\mathrm{mL}$ glass bottle i…

Problem 4 Hard Difficulty

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