# "Universal" gas constant(/s)

• May 4th 2009, 02:23 PM
TNB
"Universal" Gas Constant(/s)
I'm having a bit of trouble with the ideal gas law's universal constant(/s). If it's so universal, then why are there 3 numbers? And when do you use each one?

On my notes, it says that the universal gas constants are .0821 (labled "atm"), 62.4 (labeled "moles") and 8.314, which is not labeled at all.
I'm guessing that each number is used when trying to find the amount of atms, moles or whatever it may be. and What's the third no. for? The temperature in Kelvin?

P.S. I need to study for a chemistry test/quiz that my teacher is going to give us (The General Chemistry Class) tomorrow. If someone could give me some help/review and problems dealing with Charles' law, Boyle's law, Gay-Lussac's Law (My teacher pronounces the name Gay-Lou-Sack. Is this correct?Or is it more like :"Gie lussahk"?) The Combined Gas Law, the Ideal Gas Law, and Dalton's Law of Partial Pressures, that would be great. If you don't want to do the review, then just throw me some problems. (Save for the Ideal Gas Law- I need help with that.)
• May 4th 2009, 02:26 PM
e^(i*pi)
Quote:

Originally Posted by TNB
I'm having a bit of trouble with the ideal gas law's universal constant(/s). If it's so universal, then why are there 3 numbers? And when do you use each one?

On my notes, it says that the universal gas constants are .0821 (labled "atm"), 62.4 (labeled "moles") and 8.314, which is not labeled at all.
I'm guessing that each number is used when trying to find the amount of atms, moles or whatever it may be. and What's the third no. for? The temperature in Kelvin?

They are the same thing represented in different units. Which value you pick depends on what units the variables in your equation have.

R= $8.314\: JK^{-1}mol^{-1}$

For example, if you're using the ideal gas equation $pV=nRT$ then R would be 8.314 if n was moles and T was in kelvin.

Conceptually it's the same as saying 2+2=3+1 = 16/4 = 4
• May 4th 2009, 02:36 PM
TNB
So it just deals with what you're trying to find?

EDIT: Hey, I remember you!
• May 4th 2009, 03:01 PM
e^(i*pi)
Quote:

Originally Posted by TNB
So it just deals with what you're trying to find?

EDIT: Hey, I remember you!

Thanks XD

The value of R is a constant but it's just expressed in different units depending on what else is in the equation. They've just converted the units and they'll all have the same dimensions
• May 4th 2009, 03:42 PM
TNB
Is that a yes?
Show me some examples.
Heh.. Just got reminded of Super Smash Bros Brawl/ Captain Falcon- "Show me your moves!"

(Punch)-FAALCOONE... PAAAWNCH!
• May 5th 2009, 03:13 AM
fardeen_gen
Physical significance of $R$ and its values in different units:

For 1 mole of an ideal gas,
$R = \frac{PV}{T} = \frac{\mbox {Pressure}\times\mbox {Volume}}{\mbox {Temperature}} = \frac{(\frac{\mbox {Force}}{\mbox {Area}}) \times (\mbox {Area}\times\mbox {Length})}{\mbox {Temperature}} = \frac{\mbox {Force} \times \mbox {Length}}{\mbox {Temperature}}$

Thus, $R$represents work done per degree per mol. The values of $R$ may be calculated in different units.

Since, 1 mole of any gas at NTP ( $273\ K\ \mbox {and}\ 1\ atm$) occupies $22.4\ \mbox {litre}$.

$R = \frac{1 \times 22.4}{273} = 0.0821\ \mbox {lit-atm per degree per mol}$

To calculate $R$ in CGS unit,
$n = 1\ mole$
$P = 1\ atm = 76\ cm\ Hg = 76 \times 13.6 \times 981\ dyne/cm^2$
$V = 22400\ cc$
$T = 273\ K$
$
R = \frac{76 \times 13.6\times 981\times 22400}{273} = 8.314 \times 10^7\ erg\ degree^{-1}\ mol^{-1}
$

While solving the numericals one should keep in mind proper units.
• May 5th 2009, 03:18 AM
fardeen_gen
Try this numerical:
Two closed vessels of equal volumes which contain air at 105 kPa and 300 K are connected through a narrow tube of negligible volume. If one of the vessel is maintained at 300 K and other at 400 K, what will be the new pressure in vessel? Also calculate ratio of number of moles in each vessel.
• May 7th 2009, 02:38 PM
TNB
Uh, did you mean each vessel?
On the part that says "what will be the new pressure in vessel?"
• May 7th 2009, 09:16 PM
fardeen_gen
Quote:

Originally Posted by TNB
Uh, did you mean each vessel?
On the part that says "what will be the new pressure in vessel?"

Both the vessels are connected through a narrow tube. So pressure in each vessel is same. You are meant to find new pressure of the system(consisting of two vessels connected to each other which is equivalent to a single vessel).