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Flask contains 1OOL of helium gas at 200 torr. Flask 0.500 atm: Flask contains 2.00L of contains OOL ofneon gas at by tubing and stopcocks argon gas at 400 torr. T...

Question

Flask contains 1OOL of helium gas at 200 torr. Flask 0.500 atm: Flask contains 2.00L of contains OOL ofneon gas at by tubing and stopcocks argon gas at 400 torr. The` prevent the gases- flasks are from mixlng: connected opened. Assuming the volume of the = The stopcocks are then the system? tubing negligible; what is the total pressure

Flask contains 1OOL of helium gas at 200 torr. Flask 0.500 atm: Flask contains 2.00L of contains OOL ofneon gas at by tubing and stopcocks argon gas at 400 torr. The` prevent the gases- flasks are from mixlng: connected opened. Assuming the volume of the = The stopcocks are then the system? tubing negligible; what is the total pressure



Answers

A 275 -mL flask contains pure helium at a pressure of 752 torr. A second flask with a volume of 475 mL contains pure argon at a pressure of 722 torr. If we connect the two flasks through a stopcock and we open the stopcock, what is the partial pressure of each gas and the total pressure?

So in this problem, we have to Gas is helium and are gone. And we have the initial pressures and volumes of each of them. And we have the final total volume. And we want to find the partial pressure of each of these. And so, to do this, we can use the relation P one V one equals p to V two for each of these gases separately. And so for helium, Rp one is seven fifty to tour. Our V one is two seventy five milliliters, Rp two is unknown and R V two is seven fifty milliliters. That's the total, the new total volume. And so we can rearrange this equation above to solve for pee too soapy one V one over V two. And when we plug in these values, we get that P two is two seventy five point seven tour and now we can do the same thing for are gone. The P one of argon is seven twenty to tour the V one of argon is for seventy five milliliters. Again, the P two is our unknown and the V two is seven fifty mil leaders as it is with the helium. Since that's now the final volume for both. And we can use the same equation on the left to find that Rp two equals four. Fifty seven point three tour. And so now we want to find the pea total, which is equal to the partial pressure of helium, plus the partial pressure of argon. And we just found those on the previous page. The partial pressure of helium is to seventy five point seven tour, and the partial pressure of argon is for fifty seven point three tour. And so our p total. We just add those together and we have seven thirty three tour and so this is the final pressure.

Solving party of this problem. So here I can write the value P two is equal to B one B one by P two which is equal to 6 18. Multiplication four letter by four plus two later which is equal to 4 80 M. As the partial pressure of energy in the mixture. Now calculating the value of P. Two and two which is equal to 3 80 M. Multiplication two later 3 80 M. Multiplication to later by B 284 plus too little which is equal to 1 80 M. As the partial pressure of and to now solving part B. So here I can write live at your P. Total is equal to blog plus P and two which is required to four plus one, which is equal to 5 80 M. Now solving part C of this problem. So here I can write the value of X at G is equal to 4 80 M by four plus 1 80 M. Which is equal to 0.8 as the mole fraction of William

Solving party of this problem. So here I can write develop Even is equal to B two B two by P. Even which is required to 2.40 80. M. Multiplication given plus five liter by 3.25 p a t. M. So having it further, I can write the value of even by even plus five tweeted is equal to 2.40 by 3.25 18. And for the simplification I can devalue three points 25 P one is equal to 2.40 P. Even plus 2.40 multiplication by pleated, going forward and simplifying it for them. So I can write the value even the report 2.40 multiplication, five ft 10 by 3.25 minus 2.40 which is equal to 14.1 litre as the volume of the first container. I hope you understand it.

This question. Max, What is the total gas pressure for a steal flask that contains only oxygen in water? So oxygen and a partial pressure of Sierra Point for one indium and water vapor at personal pressure of 0.588 So in order to calculate the total gas pressure, we need Thio add together the two different personal pressures. We need the personal pressure, oxygen in partial pressure of water people. If there were other gases in this flask, we would also add them together. In this case, this question is a fairly straight board. It just has to. So if we go back to question the partial pressure of oxygen, his zero for one and we're gonna add that into the personal pressure of water vapor, which is your own point Bye, Jake, it's now it's just tradition. Question. It was one of value. Fight was for his nine. Okay, zero. So it 0.99 a t. And that's your answer. Simple is that adding the two different personal pressures together to get the total pressure. Okay,


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