Specialty Chemistry Forums > Citizen Chemist
Musings on Ideal Gas Law PV=nRT
billnotgatez:
As a citizen scientist I like to look at problems involving the Ideal Gas Law. I treat them similar to doing crossword puzzles. So I decided to create a thread with my musings on the subject. Anyone can post here, but please do not post your homework problems. You can do that in a more appropriate board and get better response.
billnotgatez:
Please correct me if these thoughts have errors ---
When analyzing some problems you can treat them as an ideal gas in situation 1 and situation 2.
Given the general Ideal Gas Law formula
PV=nRT
Both situations can be represented mathematically by
(using dot for multiplication and slash for division)
Situation 1 ---- (P1 . V1) = (n1 . R . T1)
Situation 2 ---- (P2 . V2) = (n2 . R . T2)
It follows mathematically
Situation 1 ---- (R) = (P1 . V1) / (n1 . T1)
Situation 2 ---- (R) = (P2 . V2) / (n2 . T2)
Since R is the same in both situations (it is a constant)
(P1 . V1) / (n1 . T1) = (P2 . V2) / (n2 . T2)
Algebraically this can be converted to (cross-multiplying)
(P1 . V1 . n2 . T2) = (P2 . V2 . n1 . T1)
Therefore rearranging the variables
(P2) = (P1 . V1 . n2 .T2) / (V2 . n1 . T1)
Or
(V2) = (P1 . V1 . n2 . T2) / (P2 . n1 . T1)
Or
(T2) = (P2 . V2 . n1 . T1) / (P1 . V1 . n2)
Or
(n2) = (P2 . V2 . n1 . T1) / (P1 . V1 . T2)
billnotgatez:
I hate significant digits!!!!!!!
One source has the Ideal Gas Constant as
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. The relative uncertainty is 1.8 × 10-6.]
Another source has the volume of an Ideal Gas at Standard Temperature and Pressure
22.414 L
Another source has the Standard Temperature at
273.15 K
So using PV=nRT I thought I would derive either the constant or the volume for one mole of Ideal gas at Standard Temperature and Pressure
Given
Formula
PV=nRT
Standard Temperature
273.15 K
Standard Pressure
1 atm
Volume
22.414 L
Number of molecules (mole)
1 mol
Ideal Gas Constant
0.08205746(14) L atm K-1 mol-1
Using dots for multiplication and slashes for division
(P . V) = (n . R . T)
(1 atm . 22.414 L) = (1 mol . 0.08205746(14) L atm K-1 mol-1 . 273.15 K)
Or
solving for volume
(V) = (n . R . T) / (P)
(22.414 L) = (1 mol . 0.08205746(14) L atm K-1 mol-1 . 273.15 K) / (1 atm)
Or
solving for the constant
(R) = (P . V) / (n . T)
(0.08205746(14) L atm K-1 mol-1) = (1 atm . 22.414 L) / (1 mol . 273.15 K)
The result for the volume solving is
(22.414 L) = 22.413995199
which upon visual inspection seems very close
The result of the constant solving is
( 0.08205746(14) L atm K-1 mol-1) =
0.082057477576423210690097016291415
which on visual inspection seems very ugly
but remember that (14) is a standard deviation
so the constant could be from 0.08205732 to 0.08205760 for one standard deviation.
still that is ugly
In any case both results are very close
---------------------------------
Now let us simplify the measurements
Given
Formula
PV=nRT
Standard Temperature
273 K
Standard Pressure
1 atm
Volume
22.4 L
Number of molecules (mole)
1 mol
Ideal Gas Constant
0.08205 L atm K-1 mol-1
Using dots for multiplication and slashes for division
(P . V) = (n . R . T)
(1 atm . 22.4 L) = (1 mol . 0.08205 L atm K-1 mol-1 . 273 K)
Or
solving for volume
(V) = (n . R . T) / (P)
(22.4 L) = (1 mol . 0.08205 L atm K-1 mol-1 . 273 K) / (1 atm)
Or
solving for the constant
(R) = (P . V) / (n . T)
(0.08205 L atm K-1 mol-1) = (1 atm . 22.4 L) / (1 mol . 273 K)
The result for the volume solving is
(22.4 L) = 22.39965 (rounds to 22.4)
which again upon visual inspection seems very close
The result of the constant solving is
( 0.08205 L atm K-1 mol-1) =
0.082051282051282051282051282051282
(rounds to 0.08205)
which is not so ugly
--------------------------------------------------------------------
Now i notice that some people use 0.0821 for the ideal gas constant
Given
Formula
PV=nRT
Standard Temperature
273 K
Standard Pressure
1 atm
Volume
22.4 L
Number of molecules (mole)
1 mol
Ideal Gas Constant
0.0821 L atm K-1 mol-1
Using dots for multiplication and slashes for division
(P . V) = (n . R . T)
(1 atm . 22.4 L) = (1 mol . 0.0821 L atm K-1 mol-1 . 273 K)
Or
solving for volume
(V) = (n . R . T) / (P)
(22.4 L) = (1 mol . 0.0821 L atm K-1 mol-1 . 273 K) / (1 atm)
Or
solving for the constant
(R) = (P . V) / (n . T)
(0.0821 L atm K-1 mol-1) = (1 atm . 22.4 L) / (1 mol . 273 K)
The result for the volume solving is
(22.4 L) = 22.4133 (rounds to 22.4)
which again upon visual inspection seems very close
The result of the constant solving is
( 0.0821 L atm K-1 mol-1) =
0.082051282051282051282051282051282
(rounds to 0.0821)
which is only slightly ugly
-------------------------------------------------------------------
I guess you have to take the lumps when you divide
billnotgatez:
See previous post this thread for the derivation an ideal gas in two situations.
http://www.chemicalforums.com/index.php?topic=40591.msg155048#msg155048
One can use the equation
(P1 . V1) / (n1 . T1) = (P2 . V2) / (n2 . T2)
to deduce various answers to problems involving ideal gasses.
For instance if you are given the beginning temperature and pressure plus the final temperature you can deduce the final pressure when looking at an ideal gas in a closed non-expandable container.
Since the container is closed and will not expand, you know that the volume and the number of molecules remain constant.
Algebraically then the n1 and n2 cancel out as well as the V1 and V2, which leaves the following formula.
(P1) / (T1) = (P2) / (T2)
Solving for P2
(P2) = (P1 . T2) / (T1)
Remember:
The temperature used in the equation of state is an absolute temperature: in the SI system of units, Kelvin; in the Imperial system, degrees Rankine.
Pressure should be absolute as well, for instance Absolute Pounds per Square Inch in Imperial system.
zaphraud:
Problems that are NOT homework that involve this?
I recently used this to make a guesstimate as to how many helium balloons a friend would need to float a cupcake at a birthday event. The answer is a whole lot more than humans (including myself, I noticed) usually expect using nothing but hunches.
Apparently there is something in our psychology that causes us to dramatically overestimate the lifting power of a balloon when we just think about how much it feels like it could do. :)
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