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Container A holds 722 mL of ideal gas at 2.50 atm. Container B holds 19 mL of ideal gas at 4.40 atm: If the gases are allowed to mix together; what is the resulting...

Question

Container A holds 722 mL of ideal gas at 2.50 atm. Container B holds 19 mL of ideal gas at 4.40 atm: If the gases are allowed to mix together; what is the resulting pressure?Numberatm

Container A holds 722 mL of ideal gas at 2.50 atm. Container B holds 19 mL of ideal gas at 4.40 atm: If the gases are allowed to mix together; what is the resulting pressure? Number atm



Answers

One litre of gas $\mathrm{A}$ at 2 atm pressure at $27^{\circ} \mathrm{C}$ and two litres of gas $\mathrm{B}$ at $3 \mathrm{~atm}$ pressure at $127^{\circ} \mathrm{C}$ are mixed in a 4 litre vessel. The temperature of the mixture is maintained at $327^{\circ} \mathrm{C}$. What is the total pressure of the gaseous mixture? (a) $3.93 \mathrm{~atm}$ (b) $3.25 \mathrm{~atm}$ (c) $4.25 \mathrm{~atm}$ (d) $6.25 \mathrm{~atm}$

So for this question, we have the following information given we will start with the ideal gas law P. B equals N. R. T. And we are asked to find the pressure for this gas in atmosphere. So first I will just solve this equation for peace api calls and our T over V. And we're going to substitute the following values over here. So N is too R is 0.08-1 T. is 546 Calvin and we is 44.8 L. I'm not going to put the units because I know they will cancel out based on the R value that I used. So let's just block that in our calculator two times 0.08-1 times 546, divided by 44.8. That is going to be to atmosphere. So the pressure for this gas is going to be to a T. M.

First thing that we need to do is figure out the moles of gas. They're already present in the container. That is at 28.2 atmospheres will rearrange the ideal gas law. Take our pressure volume are and then temperature 26 plus 2 273.15 To get Calvin, we see we have 61.69 moles of gases present. In order to increase the pressure to 75 atmospheres. We don't need to figure out the total moles of gas that need to be present. So we'll do the exact same calculation. But now with 75 atmospheres, you get 100 64.6 Moles are required. So if we need 170 I'm sorry. 164.6 moles. We already have 61.69 The difference will be the moles of me on that we need. We can then take the moles of neon as more mass to get the massive me on that we need to add just over 2000 grams neon about 2066 grounds

So this is problem 59 and the question asks, I make sure three gases A, B and C is a total pressure off 6.11 atmospheres. The partial pressure off gas A is 1.68 atmospheres out of gas B is 3.89 atmospheres. What is the prior show? Pressure of gassy. So So we know so far that the total pressure total partial pressure IHS 6.11 atmospheres. Then we know partial pressure of A which is 1.68 atmospheres are we know partial pressure of B which is three point 89 atmospheres. So by reading this problem ah, you can already figure out that this is just simple mats. So the total partial pressure consists off partial pressure of hey, partial pressure be pressure pressure off. See? So the mixture off three gases on, we need to find partial pressure. See, So we can just We arranged this equation for partial pressure off gas. See which will look like this. So we just need Toto off all the partial pressures. I mean to minus apply pressure, pressure A a minus bi harsh pressure be no, and he just put the values. And so the total partial pressure is 6.11 atmospheres. Then Parcher pressure survey is 1.68 atmospheres and pressure pressure be waas 3.80 nine at was fierce. Um, the answer for this problem is 0.54 atmospheres, and that's it.

So this problem gives us a certain mass of oxygen and a certain massive carbon dioxide, both Ingram's. It tells us the temperature, and it also tells us the total pressure an A T. M. And it's wanting us to find the partial pressure of oxygen and the partial pressure of carbon dioxide in a team. So to do this, we need to know how to find partial pressures in general. So partial pressures is equal to it's small fraction times the total pressure of the system. So it's more fraction it's going to be equal to the moles of oxygen divided by the total moles. So we need to find that. So we need to use our more masses to get these values and tools. And then we know the molar mass of carbon dioxide is 44 grams per one move, so we know the individual amount of moles. But we also need to know the total number of moles cause we need to use that for the mole fraction. So our total number of moles is equal to 3.119 moles. So now we can find the partial pressure. So the evil infection of 02 is 1.64 divided by the total number of moles. Times the total pressure which the problem gave us 9.21 a team and then we get four point 85 eight year and then the partial pressure of CO two. So we put again the moles of CO two and then we divide that by the total number of moles multiplied by the total pressure and we get 4.35 80 m and those are the partial pressures of Oxygen and Co two.


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