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Topic: Inversion in gas law  (Read 2513 times)

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Offline Elric82

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Inversion in gas law
« on: June 20, 2018, 06:41:08 PM »
So, I’m not in the university yet, but have been studying chem on my own in preparation. For a long time. For some reason I skipped the section in my text that deals with the gas laws. My question deals with the combined gas law, P1•V1/T1 = P2•V2/T2. Now, let’s say you want V2, and it becomes V2 = V1 • P1/P2 • T2/T1. How did they arive at this new formula? I don’t think I’m asking to be spoon fed, but I feel I’m missing something simple. I am in the text “ Chemistry, an introduction to general, organic, and biological chemistry, 10th edition” by Karen C. Timberlake.

I don’t understand why T1 and T2 have been inverted. I’ve also read to forget the combined gas law, and all the rest as well, and just use the ideal gas law. Like I said, I know I must be missing something simple.

Any input is surely appreciated.

Offline Arkcon

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Re: Inversion in gas law
« Reply #1 on: June 20, 2018, 07:20:30 PM »
Treat it like algebra.  Start with P1*VI/T1-=P2*V2/T2 and convert to the one you want by multiplying and dividing from both sides until, you get the formula you want.  And you'll see why things flip.

Then, apply it to the real world: Say You have a balloon at a certain pressure volume and temperature.  Then you make it cold.  Where would its pressure and volume go?
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Offline Elric82

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Re: Inversion in gas law
« Reply #2 on: June 21, 2018, 06:37:16 PM »
Well, that’s the problem. I mean with you’re balloon analogy. My mind knows instinctively what happens to the balloon. It will lose  volume. But, according to Boyle’s law, the pressure in the balloon will increase, because P and V are inversely proportionate.

I found the answer to the problem, but I found it through trial and error. I will state the problem, and give my answer. My answer was derived by the fact that I knew that as the pressure decreased by a value of four, then the volume must increase by four. This turned out not to be exactly true. I figured this as an inverse property. And then there’s the temp thing. Gay-Lussacs law says as the temp increases, the pressure increases, which makes sense. But that’s not what happens in this particular question. So, I will quote the question verbatim from the text and give my answer.

A 25.0 ml bubble is released from a divers air tank at a pressure of 4.00 atm and a temp of 11*C. What is the volume (ml) of the buble when it reaches the ocean surface, where the pressure is 1.00 atm and the temp is 18*C?

So, I took the change in P1 & P2 as a value of 4, and knew that V2 was close to 100 ml.
Actually I thought it was 100 ml. But, when I ran it through the original equation (combined gas law) the numbers were close, but not the same.

My answer for V2 was 102.464789. In other words 102. But that took quite a bit of calc work to derive; I guess maybee because my algebra is a little rusty. I’m actually better in the back of the textbook than I am here, and was gonna skip this stuff, but I know it’s gonna come up at the university, if I ever make it there.

Sorry for the long post, but you’re response made me open my eyes. I’ve realized that I’m gonna have to learn up to Calculus 3 for the degree I want, but I don’t see that happening on my own. I’m trying to teach myself everything I can on my own, and thought I was doing good until now

I still don’t understand how to manipulate the equation to come up with the second equation.  Like I said, I found the answer through trial & error. Maybe I need to dig my algebra tombs out and dust them off a little. I know this is basic stuff here.

Offline Borek

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Re: Inversion in gas law
« Reply #3 on: June 22, 2018, 02:59:50 AM »
I still don’t understand how to manipulate the equation to come up with the second equation.  Like I said, I found the answer through trial & error. Maybe I need to dig my algebra tombs out and dust them off a little. I know this is basic stuff here.

Definitely brush up the algebra, solving the combined gas law for any of the variables is a trivial stuff, so if you have problems on this level you won't be able to move ahead with anything else.
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Offline mjc123

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Re: Inversion in gas law
« Reply #4 on: June 22, 2018, 04:52:39 AM »
Quote
But, according to Boyle’s law, the pressure in the balloon will increase, because P and V are inversely proportionate.
That is only true at constant temperature. If both pressure and temperature are varying, you have to account for both. If it makes it easier, you can do it in two steps - a change in pressure at constant T, followed by a change in T at constant P (or the other way round).

Quote
So, I took the change in P1 & P2 as a value of 4, and knew that V2 was close to 100 ml.
That is correct as a first step (accounting for the change of pressure) [But express it more precisely; P decreases by a factor of 4, so at constant T, V increases by a factor of 4.]. But you have a temperature change to account for as well. At constant pressure, volume increases with temperature. The temperature increases from 11 to 18°C (284 to 291 K). So if you multiply your value of 100 ml by 291/284, you get the answer, 102.46...

Offline billnotgatez

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Re: Inversion in gas law
« Reply #5 on: June 22, 2018, 06:20:24 AM »
I do not know if this will confuse you but here is some algebra involving the ideal gas law.
http://www.chemicalforums.com/index.php?topic=40591.msg155048#msg155048

There are many posts involving the ideal gas law on this forum and you can find them using the forums search feature (see menu top of page).

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