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A welder using a tank of volume7.90×10−2 m3m3 fills it with oxygen (with amolar mass of 32.0 g/molg/mol ) at a gauge pressure of3.15×105 PaPa and temp...

Question

A welder using a tank of volume7.90×10−2 m3m3 fills it with oxygen (with amolar mass of 32.0 g/molg/mol ) at a gauge pressure of3.15×105 PaPa and temperature of37.0 ∘C∘C. The tank has a small leak, and in time some of theoxygen leaks out. On a day when the temperature is 21.0 ∘C∘C,the gauge pressure of the oxygen in the tank is1.90×105 PaPa .Part A: Find the initial mass of oxygen.Part B:Find the mass of oxygen that has leaked out.

A welder using a tank of volume 7.90×10−2 m3m3 fills it with oxygen (with a molar mass of 32.0 g/molg/mol ) at a gauge pressure of 3.15×105 PaPa and temperature of 37.0 ∘C∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.0 ∘C∘C, the gauge pressure of the oxygen in the tank is 1.90×105 PaPa . Part A: Find the initial mass of oxygen. Part B: Find the mass of oxygen that has leaked out.



Answers

A welder using a tank of volume 0.0750 $\mathrm{m}^{3}$ fills it with oxygen (molar mass 32.0 $\mathrm{g} / \mathrm{mol} )$ at a gauge pressure of 3.00 $\mathrm{x}$ $10^{5} \mathrm{Pa}$ and temperature of $37.0^{\circ} \mathrm{C} .$ The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is $22.0^{\circ} \mathrm{C},$ the gauge pressure of the oxygen in the tank is $1.80 \times 10^{5} \mathrm{Pa} .$ Find (a) the initial mass of oxygen and (b) the
mass of oxygen that has leaked out.

All right. So want to find the mass, which is a small M, and that's equal to the number of most times the molar mass who have most of what you want to find in both cases. So let's get started. So part A using PV equals NRT number of moles is just p V over r T. This is the ideal gas law pressure, which has given us gauge pressure, should be converted to absolute pressure. So three times five plus one point out. 1, 10 10 and five So three plus 1.1. This is atmospheric pressure, right? Sometimes 10 of the five Pascal's is multiplied by volume, which is 0.75 cubic meter on divided by gas constant 8.315 Jules from or types Calvin Times temperature, which is to 73 plus, um, plus 30 uh, seven uh degrees Celsius. So this please 7 to 37. So in Calvin's, that's 310 you get get 11.67 malls for the first part. So that's the initial number of moles. Um, and so therefore am will be 10 times AM big. Um, so this is 11.67 moles times smaller mass of oxygen, which is 32 g promotes, and you've got 374 g for the inertial mass. It's called the M one. Okay, uh, I'm one initial masses 374 party. Um, first thing to do is just repeat this thing's process, but for final, and so we're going to use the subscript two final number of moles is to be to over our t tune. So everything to your screens final. Um, now you have final pressure of 1.8 10 10 5 gauge pressure. So 1.8 plus 1.12 point 81 times 10 to the five past cows is the final pressure times, mhm times. The volume, um, remains the same. So that's 0.75 cubic meters in dividing this by gas constant again times a temperature which is 2 73 plus 20 to send to 95 Calvin. And so your final number of most is eight point six bulls. All right, so your final mass is and two times answer. It's 8.6 times 32 grand per mole, which gives you 2. 70 for cramps. However, this is not your final answer because you want to know the leaked Mass. So M l leaked mass that's equal to initial minus finals. That's your remaining mass. Right on DSO initially in 3 74 final last 2 74 leg gives you 100 grands for your final, uh, answer for lead to Mass and part B.

So we have our givens. They're giving us the volume point zero seven five zero meters cubed and then we have a Mueller Mass. Of course of oven oxygen molecule being thirty two grams Permal. We have a date gauge pressure equaling three times ten to the fifth. Again, this is the gauge pressure and this isn't Pascal's. Then we have a temperature of thirty seven degrees Celsius. This is equaling three hundred and ten Calvin were simply adding to seventy three to the temperature and Celsius toe to convert to Kelvin. And then we have process where the volume is not changing. The Mueller Mass, of course, is not changing. But however, we are changing pressure. We're lowering pressure toe one point eight times ten to the fifth Pascal's and then the temperature will be now twenty two degrees Celsius or two seventy three to two seventy three plus two, twenty two two ninety five Calvin. So temperatures being lowered, pressure's being lowered. And it's saying, what was the initial mass of the what was the initial mass of the oxygen and how much oxygen was lost after this process here. So to find the initial mass, the initial massive oxygen. We would thought we could say PV equals and Artie used the IO cast law and we can say that the number of moles and is going to be equal to the mass of the whatever's inside the system, divided by the molar mass of the molecules within the system. In this case, we're going to have oxygen molecules, so that be the more massive O two. We can then rewrite to this equations and say that PV equals and Marty over the more mass and then we can solve for ahh, lower case M the mass. We can say that this is going to be PVM over Artie. And at this point, we can solve so we can say equals. And then we would have we would start by using our pressure here. Now, this pressure here is the gauge pressure. So the pressure here, plus the atmospheric pressure. So right here, um, they gave us the gauge pressure. However, we need Teo in order to use this equation. Me to find the total pressure. So it would be that pressure right here, plus the atmospheric pressure, the atmospheric pressure being, of course, one hundred and one thousand three hundred twenty five Pascal's So at that point, we can say Okay if we add this value here three times ten to the fifth plus one, one o one, three to five past cows, we can say that there's going to be four point zero one three times ten to the fifth. Pascal's time's our volume of point zero seven five zero meters Cute and then we have named Mueller Mass of oxygen thirty two grams per mall and again, we're using grams here, so we know that the mass is going to be in grams. Just keep track of what you and it's here you're using. We're going to divide by on. We're using Pascal's and Meters Cube, so we'LL use eight point three one four for the ideal gas constant. And then we'LL have three hundred and ten Calvin our initial temperature. We find out there's going to be approximately three seventy three point six nine grams, so this would be our initial mass. Now it's saying thiss process right here is occurring. What happens? How much oxygen leaks out? So it's got a new board. We have, um, equals again. We're going to use that same exact formula PVM over Artie. Except that now we're going to be using the initial d. Sorry, the final pressure. So again it will be this pressure plus the atmospheric pressure. This again is a gauge pressure. So in order to find the pressure that we use in the formula we had, we just simply have to add the atmospheric pressure to that. That was going to be two point eight one three times ten to the fifth. And then we have our volume. That does not change point zero seven five zero meters cube and then times the Mueller mass of oxygen again. We're using grams, so we know that the answer will be in grams and then we're using at Miss Rather using Pascal's and meters cubed. So our r should be again eight point three one four times the final temperature of two hundred and ninety five. Calvin and our mass is going to be two seventy five point two six grams Now. This is not our frontal answer. Of course we need Thio. See the mass that's lost. So Delta m, we can have the initial mass three seventy three point six nine minus two seventy five point two six. This will be equal to ninety eight point four three grams. So we'LL have ninety eight point four three grams of O two lost. That will be our answer for part B. And that's the end of the solution. Thank you for watching.

So this question belongs to the Canada three of gases in which we have given oxygen slender of initial volume 30 litre and initial gas pressure even it is equal to 15 80 M. That is equal to 15 molecular by 1.1 molecular weight. And to the power five paschal and initial temperature even equals to 27 degrees integrate that is equal to 300 Calvin. Okay. And the final values are we have given some of season guests has been withdrawn and final pressure is equal to 11 80 M. That is 11 to 1.1 more player bait. And to the power five paschal And the temperature dropped to eight equals to 17° integrated that is 2 90 Calvin. So we have to determine the mass that is taken out from the cylinder that is delta M. So and we have given molecular mass and this is equal to 30 to you. So we can calculate that from the ideal situation PV equals too N RT So and this can be written as mass river by molecular mass molecular by RT So from here, mass equation is equal to tv murder capital M. D. Barbie artie So for this condition first condition the mass initial and when this is equal to even be even and develop our even so substituting values of even which is equal to 15 molecular B 1.1 molecular weight. And to the power five We won which is 30 later. So 30 more player by 10 to the power minus three m Q. And molecular weight which is 32. Okay so and divided by 32 g. It can be taken as so are which is equal to 8.3 and four. And temperature T one which is 300 Calvin. So initial mass M one comes out to be from here. Uh this is equal to 5 84.84 g. And similarly the final mass M two. This is equal to final pressure which is this value. So 11.11 molecular bay 1.1 molecular weight. And to the power five and volume of the cylinder will remain same. So hence this is equal to 13 particular bait. And to the power minus three molecular molecular weight is also same deal whereby universal guest, constant muscular be final temperature which is 2 90 Calvin. So final mass comes out to be 4 53.1 g. So now we can calculate this change in mass that is delta M. So this is equal to one minus M two. Initial mass minus final mass. So 5 84.84 g minus 4 53.1 g. Okay, so from here the mass escaped from the Or mass released or what must be drawn from the slender comes out to be 131.74 g, or it can be taken as 0.132 program. Okay, so this become the answer for this problem. Okay. Thank you.

Okay. Use the ideal gas law. PV equals NRT to calculate the number of moles and have t r p V to calculate the first. So the volume for its elliptical container is pyre squared out right? So are as 0.45 m Anil is 1.5 meters and so that gives you volume off 0.953 Muted. Cute, uh, and so therefore the end, which is PV over our team. It's the number of balls is equal to pressure and 20 that's 21 atmospheres. The 21 atmosphere. It's times no 13 times 10 to the five Pascal's for atmosphere. Do you have that in Pascal's now times volume of We just Found this 0.953 m cute over 8.31 Jules per mole. A Calvin. That's the gas Constant Times Lot 273 loss. 207 3 plus 22. So that's to night, so it's 22 degrees Celsius and to 95 degrees. Calvin temperature gives you a number of moles and 827 in Part B. We're just using the fact that the models recalled to massive elements so Mass will be end times the molar. That's so 827 moles times 32 g So 32 times to my sweet killed rounds per mole, um, for oxygen and so get 26.5 killer grounds the Earth.


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