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Offline MN1988

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Gas Law problem
« on: October 03, 2014, 12:22:49 PM »
Hello I am having a problem with the ideal gas equation and the combined gas law. My teacher said that both can be used for this problem. He said that the combined gas law can be used if one side of the equation is in s.t.p.

The question:
Calculate the volume that 20g of oxygen gas will occupy at 25 °C and 20kPa pressure

first thing I done was turn the 20g to moles
20/16 = 1.125 moles  (divide by mmass of oxygen)

then i wanted to see what volume this would occupy at s.t.p.
1.125 x 22.4 =25.2 L

Then I changed temperature to Kelvin: 25+273 = 298 K

For the combined gas law:
P1V1/T1 = P2V2/T2
P1V1T2/T1P2 = V2
(100)(25.2)(298)/(273)(20) = V2
137.54 = V2

For the ideal gas equation:
I first converted pressure to Pa: 20 x 1000 = 20000
PV = nRT
V = nRT/P
V = (1.125)(8.31)(298)/(20000)
V = 0.13929
but I had to multiply by 1000 to change from meters cubed to L
V = 139.29 L

I have two different values for the volume, they are very close but not the same and I've seen plenty of examples where both the combined gas law and the ideal gas equation get the same result.
I've done this calculation many times and can't seem to find my flaw.
Please help

Offline sjb

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Re: Gas Law problem
« Reply #1 on: October 03, 2014, 01:30:38 PM »
Hello I am having a problem with the ideal gas equation and the combined gas law. My teacher said that both can be used for this problem. He said that the combined gas law can be used if one side of the equation is in s.t.p.

The question:
Calculate the volume that 20g of oxygen gas will occupy at 25 °C and 20kPa pressure

first thing I done was turn the 20g to moles
20/16 = 1.125 moles  (divide by mmass of oxygen)

then i wanted to see what volume this would occupy at s.t.p.
1.125 x 22.4 =25.2 L

Then I changed temperature to Kelvin: 25+273 = 298 K

For the combined gas law:
P1V1/T1 = P2V2/T2
P1V1T2/T1P2 = V2
(100)(25.2)(298)/(273)(20) = V2
137.54 = V2

For the ideal gas equation:
I first converted pressure to Pa: 20 x 1000 = 20000
PV = nRT
V = nRT/P
V = (1.125)(8.31)(298)/(20000)
V = 0.13929
but I had to multiply by 1000 to change from meters cubed to L
V = 139.29 L

I have two different values for the volume, they are very close but not the same and I've seen plenty of examples where both the combined gas law and the ideal gas equation get the same result.
I've done this calculation many times and can't seem to find my flaw.
Please help

What is stp? Or, under what conditions does a mole of gas take up 24litres?

Offline mjc123

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Re: Gas Law problem
« Reply #2 on: October 04, 2014, 02:31:28 PM »
Also, what is the molar mass of oxygen gas (O2)?

Offline MN1988

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Re: Gas Law problem
« Reply #3 on: October 04, 2014, 09:31:19 PM »
Thank you so much mjc123 for pointing out the molar mass of oxygen gas - I feel like an idiot for not seeing that it was a diatomic molecule.
Correction
20/32 = 0.625 moles

sjb I'm afraid I don't see what you're trying to lead me on to. Here's my logic:
s.t.p = standard temperature and pressure. (273 Kelvin and 100 kPa)
One mole of any gas takes up 22.4 L under s.t.p.
So 0.625 moles of a gas is 22.4 x 0.625 = 14 L

I've put this into the combined gas law and got 76.41 L
and when I put it the new value for n into the ideal gas equation I got 77.39 L

I know its a very small difference but I just want to be sure that I'm doing it right. Am I mixing up something about standard temperature and pressure?

Offline Borek

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Re: Gas Law problem
« Reply #4 on: October 05, 2014, 03:36:32 AM »
There are many different definitions of STP.

Ideal gas at 0°C and 1 atm has a volume of 22.414 L

Ideal gas at 0°C and 100 kPa has a volume of 22.711 L

Looks like you used 0°C and 100 kPa and 22.414 L, which is clearly wrong.
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Offline MN1988

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Re: Gas Law problem
« Reply #5 on: October 05, 2014, 07:02:41 PM »
I used 100 kPa because pressure in the question was given in kPa. I know I'm wrong - I'm looking for the path to being right.
the definition of s.t.p. i always used and was given to me is standard temperature is 0°C or 273 K
and standard pressure is 100 kPa
I know 1 atm is very close to 100 kPa. But isn't the atmosphere unit a constant taken as the average pressure at sea level and not the standardized pressure accepted by IUPAC?

Offline Borek

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Re: Gas Law problem
« Reply #6 on: October 06, 2014, 02:55:47 AM »
But isn't the atmosphere unit a constant taken as the average pressure at sea level and not the standardized pressure accepted by IUPAC?

That's more or less how it was defined initially, but it doesn't matter here.

Actually I am not sure what is your problem now, I thought my explanation clearly shows what you did wrong.

You stated

Quote
One mole of any gas takes up 22.4 L under s.t.p.

and you used this information to calculate volume of the gas. That is OK as long as the pressure is 1 atm, but then you calculated volume using ideal gas equation and pressure of 100 kPa, and you were surprised the result is different. It must be different, as these volumes reflect different pressures.
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Offline MN1988

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Re: Gas Law problem
« Reply #7 on: October 07, 2014, 11:14:07 AM »
I beg your pardon I must have misread your post - I read it again there and it makes sense.

I done it out again with the new value of 22.711. By the combined gas law I am still getting an answer of 76.41 L and by the ideal gas equation I am getting an answer of 77.39 L

At this stage I accept that they are close enough and I am very grateful for every ones help - my understand has greatly increased. Thank you very much.

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