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Topic: Verifying calculations for volume.  (Read 9656 times)

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Offline SimoneMeyer

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Verifying calculations for volume.
« on: September 05, 2009, 08:22:42 PM »
The question asks:

Natural gas is stored in an underground reservoir at a pressure of 425 kPa and at a temperature of 15 degrees Celsius. The volume of the reservoir is 1500 L. What will be the volume occupied by the gas if it's transferred to a pipeline where the temperature is at 35 degrees Celsius and 105 kPa?

This is my answer, but would like to know if it's correct.



Thank you!

Offline Borek

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Re: Verifying calculations for volume.
« Reply #1 on: September 06, 2009, 03:35:42 AM »
No, your answer is not correct. Your approach is not completely off, but try to do it more systematically, starting with equation describing the situation (one of the versions of ideal gas equation). It should be obvious then what mistake you did and where.
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Offline SimoneMeyer

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Re: Verifying calculations for volume.
« Reply #2 on: September 06, 2009, 06:13:27 PM »
Was the wrong equation/formula used?

Offline MrTeo

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Re: Verifying calculations for volume.
« Reply #3 on: September 07, 2009, 03:04:04 AM »
Try to use the state equation as Borek said... the fact is that you've used no official formula  ;)
Start from pV=nRT and think what doesn't change in the the equation going from step 1 to 2...
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Offline SimoneMeyer

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Re: Verifying calculations for volume.
« Reply #4 on: September 07, 2009, 12:39:04 PM »
If the equation is pV = nRT, would it look something like this:

100kPa (V) = 1500L x 8.314472(15) J/mol x 308K

?

Also, when it comes to the constant of 8.314472(15) J/mol, what is my J and mol?

Is there another way to make the equation so it's structured like: V = etc.... ?

Offline MrTeo

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Re: Verifying calculations for volume.
« Reply #5 on: September 07, 2009, 02:28:39 PM »
Obviously there is  ;)



But I think this is more helpful to you (moles don't change, as we suppose that the pipeline doesn't leak ;D)





Try to use these information and you'll soon get something very similar to your first equation...
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Offline SimoneMeyer

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Re: Verifying calculations for volume.
« Reply #6 on: September 07, 2009, 03:06:26 PM »
I have a slight hunch that my current answer is way way off...

I did:
   
    1500L x 8.3144721(15) x 308K
v: ------------------------------
                 105 kPa

V= 5,359,271.095

Is it in ml?

Offline MrTeo

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Re: Verifying calculations for volume.
« Reply #7 on: September 07, 2009, 04:43:18 PM »
What's that (15)? If it's 15ÂșC you should convert it in K before and most important... why don't you read carefully my posts and try at least to do something of what I've suggested you? It's the easiest and the more correct approach...
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

Offline SimoneMeyer

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Re: Verifying calculations for volume.
« Reply #8 on: September 07, 2009, 05:13:27 PM »
I thought that I was following the V = nRT
                                                   -----
                                                      p

approach.

I thought that the constant (R) was supposed to be 8.3144721(15)

Offline MrTeo

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Re: Verifying calculations for volume.
« Reply #9 on: September 07, 2009, 05:27:24 PM »
But I think this is more helpful to you (moles don't change, as we suppose that the pipeline doesn't leak ;D)





Try to use these information and you'll soon get something very similar to your first equation...

Using volume there is no real approach as you still need to find out the number of moles in the first equation... Try starting from and you'll easily find your way, and sorry if I haven't been clear  ;)
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Offline Borek

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Re: Verifying calculations for volume.
« Reply #10 on: September 07, 2009, 06:16:41 PM »
What's that (15)?

8.3144721(15) means last two digits are +/-15, this is a way of giving accuracy of the constant.

But as you are correctly suggesting R is not needed to solve the question.
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Offline SimoneMeyer

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Re: Verifying calculations for volume.
« Reply #11 on: September 07, 2009, 07:21:26 PM »
I'm really sorry, and I know that I must seem as dumb as a brick wall, but I don't understand how n1 = n2 will solve my question if I'm looking for the volume.

And if I do it without the R, would the equation be:

        pV
n = -------
         T

But am I able to just take the R out like that?

Offline MrTeo

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Re: Verifying calculations for volume.
« Reply #12 on: September 08, 2009, 01:53:49 AM »

8.3144721(15) means last two digits are +/-15, this is a way of giving accuracy of the constant.

But as you are correctly suggesting R is not needed to solve the question.

Oh I see... well, thanks Borek, never seen something like that before ^^

I'm really sorry, and I know that I must seem as dumb as a brick wall, but I don't understand how n1 = n2 will solve my question if I'm looking for the volume.

And if I do it without the R, would the equation be:

        pV
n = -------
         T

But am I able to just take the R out like that?


Ok... let's tidy up the whole thing, and don't worry if you don't understand at first glance  ;)

No one said you had to forget the R constant and write the equation without that... it's like writing a reaction without one of the products: there's something missing...
Anyway R doesn't actually need: doing some easy calculations you'll see that it's soon taken out.

As I said start from , here's what I mean:




Start from this, now it should be much more clearer
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Offline Borek

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Re: Verifying calculations for volume.
« Reply #13 on: September 08, 2009, 02:48:37 AM »

R cancels out giving the R-less equation I was talking about.
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Offline MrTeo

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Re: Verifying calculations for volume.
« Reply #14 on: September 08, 2009, 08:13:31 AM »
Just another note on the (15)... seems like it's a standard form used both by IUPAC and CODATA to express uncertainity... interesting indeed ^^
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

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