[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. 

[billnotgatez]

I continue to muse -

I ran across the below statement on another list having to do with aviation and it made me think about how I hate the term STP.

ICAO/ISO standard atmosphere is 15 C (59 F), which is more realistic than the oft-used scientific baseline of 0 C (32 F). But in either case the rate of pressure drop with altitude is approximately the same.

-----------------------

Some time ago on another thread I did a rant. Here is an edited version

<rant mode on>

All values in the ideal gas equation must be converted to absolutes. So for instance 273.15 K is the absolute and 0 C is not the absolute temperature. I can not think of a way that you can have a volume that is not absolute. But, I have had problems presented to me that the pressure was not stated as and absolute. I can remember having to convert PSI to APSI (absolute PSI) and then convert to metric or ATM. So when I see a posting having 0 ATM, I assumed that it was not absolute and had to be changed to 1 AATM  (absolute ATM)  for STP. It never occurred to me that the post was a typo. Having said this it seems every post I have seen in the past ATM is assumed to be AATM.

This is the stuff that drives me crazy since STP and ATM are something one is expected to know even though it depends on who you are dealing with when answering the question. I personally think it is poor form to use STP rather than state the temperature and pressure, and also not state pressure as absolute. Many gages do not measure absolute.

<rant mode off>

[zaphraud]

This is the stuff that drives me crazy since STP and ATM are something one is expected to know even though it depends on who you are dealing with when answering the question. I personally think it is poor form to use STP rather than state the temperature and pressure, and also not state pressure as absolute. Many gages do not measure absolute.

Yeah huh!

Turbo gauges that report "boost" are irritating in that respect; while the good ones do head negative when the driver is off the throttle, a direct report of the manifold pressure would give a much better idea how much fuel-air is being consumed and thus, contributing to power: absolute pressure has a real relationship to an engine's power output. Boost doesn't.

[billnotgatez]

Even more musing on the mathematics of the Ideal Gas Law

I have noticed that a common mistake when doing computations involving the Ideal Gas Law is failing to use the correct units. You need the correct units, if you are going to make the mathematics work properly.

The calculations require that the variables be absolute mathematics to work.

Well, for volume (V) that is easy, there is not really a negative volume. So there is an absolute volume rather than relative volume.

And, this holds true for the amount of substance (n). In this real world, you got stuff or you got nothing.

Temperature (T)on the other hand can be measured relative or absolute.
Looking at WIKI
http://en.wikipedia.org/wiki/Temperature
http://en.wikipedia.org/wiki/Celsius
http://en.wikipedia.org/wiki/Kelvin
http://en.wikipedia.org/wiki/Conversion_of_units_of_temperature

Absolute zero
Kelvin 0.00
Celsius -273.15
Fahrenheit -459.67
Rankine 0.00
Delisle 559.73
Newton -90.14
Réaumur -218.52
Rømer -135.90

Again we need to select the correct units to do the calculations. That may mean you have to convert from the temperature units you have to the absolute temperature units. Many people favor Kelvin over Rankine.

I have seen for pressure (P) measurements the old tire gauge is used. Well, most everyday gages start off at zero so that means they started with atmospheric pressure built in to the measurement. So PSIG (pounds per square inch gauge) is the same as approximately 14.7 PSIA (pounds per square inch absolute). In many of the question you get on test, the questions are biased by giving you the absolute pressure. But, sometimes the testers trick you. In any case, if you do not have absolute pressure, convert what you have to absolute.

----------
Now we have everything converted to an absolute units and then we need to select constant (R) to go with those units.
WIKI has a whole list of potential values of the constant.
http://en.wikipedia.org/wiki/Gas_constant
For instance if our units are Litre Atmosphere Kelvin Mole
0.08205746 L atm K−1 mol−1 would be used for the constant (R).

[Borek]

I have noticed that a common mistake when doing computations involving the Ideal Gas Law is failing to use the correct units.

This problem is not limited only to ideal gas calculations - my bet is that it is the most common mistake repeated on the daily basis in all schools on all continents.

Perhaps with exclusion of the Antarctica.

[billnotgatez]

I just noticed on WIKI
http://en.wikipedia.org/wiki/Gas_constant

8.3144621(75)     L  kPa K⁻¹ mol−1
8.3144621(75)    m³  Pa K⁻¹ mol⁻¹

I am now confused as to Pa and kPa having the same result for R
Maybe I need to sit down and do the math

[billnotgatez]

1m³ = 1000.0L
now the light shines on my head

[billnotgatez]

This is a rewrite of a previous post with better formatting

Please correct me if these thoughts have errors ---

When analyzing some Ideal Gas problems you can treat them as an ideal gas in situation 1 and situation 2.

Given the general Ideal Gas Law formula

[tex]    PV=nRT   [/tex]

Both situations can be represented mathematically by

[tex]    Situation 1 ~ \Longrightarrow ~~   (P_1  \times  V_1) = (n_1  \times  R  \times  T_1)   [/tex]

[tex]    Situation 2 ~ \Longrightarrow ~~   (P_2  \times  V_2) = (n_2  \times  R  \times  T_2)   [/tex]

It follows mathematically

[tex]    Situation 1 ~ \Longrightarrow ~~   (R)  =  \frac{(P_1  \times  V_1) }{ (n_1  \times  T_1)}   [/tex]

[tex]    Situation 2 ~ \Longrightarrow ~~   (R)  =  \frac{(P_2  \times  V_2) }{ (n_2  \times  T_2)}  [/tex]

Since R is the same in both situations (it is a constant)

[tex]     \frac{(P_1  \times  V_1) }{ (n_1  \times  T_1)} =  \frac{(P_2  \times  V_2) }{ (n_2  \times  T_2)}   [/tex]

Algebraically this can be converted to (cross-multiplying)

[tex]   (P_1  \times  V_1  \times  n_2  \times  T_2) = (P_2  \times  V_2  \times  n_1  \times  T_1)   [/tex]

Therefore rearranging the variables

[tex]    (P_2) =  \frac{(P_1  \times  V_1  \times  n_2  \times T_2) }{ (V_2  \times  n_1  \times  T_1)}   [/tex]

Or

[tex]    (V_2) =  \frac{ (P_1  \times  V_1  \times  n_2  \times  T_2) }{ (P_2  \times  n_1  \times  T_1)}   [/tex]

Or

[tex]    (T_2) = \frac { (P_2  \times  V_2  \times  n_1  \times  T_1) }{ (P_1  \times  V_1  \times  n_2)}   [/tex]

Or

[tex]    (n2) =   \frac{(P_2  \times  V_2  \times  n_1  \times  T_1) }{ (P_1  \times  V_1  \times  T_2)}   [/tex]