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### Topic: Compute moles of helium in a 22.4 L size Container  (Read 10109 times)

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#### ryno16

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##### Compute moles of helium in a 22.4 L size Container
« on: November 30, 2009, 11:30:08 AM »
Compute the number of moles of helium (He) gas contained in a 22.4-L-size container at 1.00 atm pressure and a temperature of 273 K.  Use 0.0821 L atm/mol K for the gas constant R.

I know you apply PV=nRT
i'm just struggling in figuring out how to apply all the numbers given in the equation to this formula.

1*22.4=n*0.821*273

is this the correct set up?

#### Borek

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #1 on: November 30, 2009, 11:46:33 AM »
Close, but you are missing 0 in the R.
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#### Wreath

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #2 on: November 30, 2009, 01:23:30 PM »
I may be ashamed later, but I thougt that 22,4 dm3 is the volume of 1 mole of any gas? so, if there is 22,4l container and 1 ATM pressure, there should be 1 mole of He exactly?

#### Borek

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #3 on: November 30, 2009, 01:49:39 PM »
I may be ashamed later, but I thougt that 22,4 dm3 is the volume of 1 mole of any gas?

Only at STP, and even that doesn't have to be true as STP is not well defined...

Quote
so, if there is 22,4l container and 1 ATM pressure, there should be 1 mole of He exactly?

Assuming temperature of 273 K - yes.
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#### Wreath

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #4 on: November 30, 2009, 02:11:01 PM »
ok, thanks.

#### billnotgatez

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #5 on: December 01, 2009, 05:09:02 AM »
This may be a sugnificant figure question since the result i get is
0.999406602329866641681501608420001

#### Borek

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #6 on: December 01, 2009, 05:51:52 AM »
Rounding errors, no doubt about it.

Note, that we don't know exact values of numbers used. They were all rounded and we were not given their accuracy. In practice that means that the last digit is somewhere between +/- 0.5 (note that decimal point refers to the digit, not the number) - so 22.4 can mean anything between 22.35 and 22.45 (more precisely 23.49999...), 0.0821 anything between 0.08205 and 0.08215 and so on. That in turn means that RT/(pV) is something between

0.08215*273.5/(0.995*22.35) = 1.01033

and

0.08205*272.5/(1.005*22.45) = 0.99097

(note: first case - cumulation of positive errors in the nominator and negative errors in denominator, second case - exactly the opposite, that guarantees maximum possible negative and positive error; this is a variant of 'crank three times' method).

We can use much better value of R & T to get smaller error, other numbers are hard to modify.
« Last Edit: December 01, 2009, 06:06:24 AM by Borek »
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#### billnotgatez

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #7 on: December 03, 2009, 09:08:10 AM »
The value for R on WIKI
0.08205746(14) L atm K-1 mol-1
The two digits in parentheses are the uncertainty (standard deviation) in the last two digits of the value
Ideal gases: 22.414 L/mol at 0 °C: 24.465 L/mol at 25 °C mols are measured n m/M n mols m mass M mass of compound.

Using R = 0.08205746 (ignoring standard deviation)
1 mol 1 atm 273.15 K we get  volume
22.413995199

#### Borek

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #8 on: December 03, 2009, 09:48:26 AM »
Using R = 0.08205746 (ignoring standard deviation)
1 mol 1 atm 273.15 K we get  volume
22.413995199

Significant digits are not the best way of expressing uncertanity, but the volume you listed is for 1.0000000000 atm, 1.0000000000 mol and 273.15000000 K...

(OK, I have not paid attention to exact number of zeros, but it should be obvious what I mean).
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#### billnotgatez

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##### Re: Compute moles of helium in a 22.4 L size Container
« Reply #9 on: December 03, 2009, 04:49:48 PM »
Yes I understand
I wonder if the person who originally posted is reading this