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Topic: The R constant  (Read 4708 times)

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Offline Archy12345

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The R constant
« on: February 11, 2014, 11:21:07 AM »
I know that the 0.082 value for R is derived from the Ideal gas law equation.

From what equation is the value 8.31J/mol K?

Offline Borek

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Re: The R constant
« Reply #1 on: February 11, 2014, 12:41:55 PM »
It is the same constant, just expressed in different units. See wikipedia page on the ideal gas constant - there is a huge table there.
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Offline Archy12345

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Re: The R constant
« Reply #2 on: February 11, 2014, 05:15:44 PM »
I am aware of the fact that they are the same constant and just expressed in different units. I was just wondering how we ended up with J/mol K from L atm/mol K.

The wikipedia page did help, thanks.

Offline Borek

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Re: The R constant
« Reply #3 on: February 11, 2014, 05:33:28 PM »
W = pΔV, volume×pressure has units of energy.
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Offline Archy12345

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Re: The R constant
« Reply #4 on: February 11, 2014, 05:43:29 PM »
Okay...I'm able to convert liter atmospheres to joules...

So then this becomes J/(mol·K).

So our ideal gas law equation then ends up being Energy=mol·R·T

Is the goal then to determine the amount of energy of a gas that is required to exert 1 atmosphere in 22.4 liters of space?

Offline Borek

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Re: The R constant
« Reply #5 on: February 11, 2014, 06:23:53 PM »
Goal was to find an equation of state for gas, and it turned out R is a necessary proportionality constant.
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Offline Archy12345

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Re: The R constant
« Reply #6 on: February 11, 2014, 06:34:43 PM »
I must not be asking my question correctly...

The 0.082 value for R comes from the ideal gas law equation at STP

(1atm · 22.4L)/(1mol · 273.15K) = 0.082atm·L/mol·K

I wish to know from which equation the value 8.314 for R is derived.

I know the units for 8.314 are J/(mol · K). These units must be a result of some equation. What is that equation?

Offline billnotgatez

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Re: The R constant
« Reply #7 on: February 11, 2014, 07:54:28 PM »
A search of WIKI gives several locations where you can read about various aspects of the Ideal Gas Law.
It seems to me that the the Ideal Gas Law evolved over time and there was a lot of empirical work done that led to various aspects of the equation to include the R constant.

Here are some WIKI entries that you may want to peruse

http://en.wikipedia.org/wiki/Equations_of_state/History
http://en.wikipedia.org/wiki/Ideal_gas_constant
http://en.wikipedia.org/wiki/Ideal_gas
http://en.wikipedia.org/wiki/Ideal_gas_law
http://en.wikipedia.org/wiki/Standard_temperature_and_pressure

Within each WIKI entry there are more links that can give you more insight.

Additionally, we have discussed the Ideal Gas Law in this forum many times.
Maybe a thread like this might help

http://www.chemicalforums.com/index.php?topic=40591.0

By the way if you do simple algebra on the Ideal Gas Law you can represent R.

R  = P.V / n.T

Then you can do many different algebraic and unit manipulations to get the various values of R.
This would include substitution of 2 of the variables in the equation with 1 variable like suggested by a previous entry in this thread discussing energy.

Can you write an equation making substitutions?



« Last Edit: February 11, 2014, 08:21:22 PM by billnotgatez »

Offline Archy12345

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Re: The R constant
« Reply #8 on: February 11, 2014, 11:49:12 PM »
Here is the answer to my question:

kPa·L = N·dm = J

101.32kPa · 22.4L = 2269.6J = 1mol · R(J/mol·K) · 273.15K

2269.6J/(1mol · 273.15K) = 8.31J/(mol·K)

Offline Borek

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Re: The R constant
« Reply #9 on: February 12, 2014, 03:01:57 AM »
I must not be asking my question correctly...

I think it is rather problem of some misunderstanding, or misuse of the word "equation".

Quote
The 0.082 value for R comes from the ideal gas law equation at STP

(1atm · 22.4L)/(1mol · 273.15K) = 0.082atm·L/mol·K

I wish to know from which equation the value 8.314 for R is derived.

From exactly the same ideal gas equation. Just because you express values using different units doesn't mean you are using a different equation.

Once you know R value using any set of units it is not a problem to convert it to another, just multiply it by conversion factor(s). For example if you have R expressed as 0.08206 atm·L/mol·K and you want to express pressure using mmHg, conversion factor is 760 mmHg/1 atm and new R value is

[tex]760 \frac {mmHg}{atm} \times 0.082057 \frac {atm \times L}{mol \times K} = 62.36 \frac {mmHg \times L}{mol \times K}[/tex]
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