[Valeder]

Hello

Today I was studying van't Hoff equation related to osmotic pressure, and I found that at the basis of this equation there is
the ideal gas law. Ok, but I don't understand the expression of the ideal gas constant, R.

I have looked on the book of Physiscs and I have found that its value is 8,318 J·mol^-1·K^-1.
On the book of chemistry I have found that its value is 0,082 L·atm·mol^-1·K^-1.

Then I have looked on the web and I have found that the ideal gas constant can be expressed in many ways, and these are two of them. But I wasn't happy, and so I tried to compare the equations of this values, but I still doesn't understand how they can express the same number with different units of measure 

Hence I saw the first equation (with joules) in basics units should be : kg ·m^2·s^-2·mol^-1·K^-1

The second one (with atm) in basics units should be : cm^3·g·cm^2·s^-2·cm^-2·mol^-1·K^-1

developing them, and removing values present in both we have:

cm^3·g=kg·m^2

I believe I do something (or even more than something  ) wrong, but I don't understand what 

I will be very happy and thankful if anybody could help me.

[Borek]

What is your pressure unit?

J·mol^-1·K^-1 = L·atm·mol^-1·K^-1
J = atm * L
work = pressure*volume
N*m = N/m² * m³
N = N

(conversion factors neglected for obvious reasons)

Note that work = pressure*volume is the same as in W = PdV

[Valeder]

What is your pressure unit?

I was using atmospheres (atm). But I remember that 1 atm is: dyne per cm²

J·mol^-1·K^-1 = L·atm·mol^-1·K^-1
J = atm * L
work = pressure*volume
N*m = N/m² * m³
N = N

(conversion factors neglected for obvious reasons)

Note that work = pressure*volume is the same as in W = PdV

Really thanks  Now I have understood the sense of it, and why they can be the same.

So if dyne per cm² is atm.
dyne is a unit of measure of cgs system, it is g*cm*s^-2, while Liter is a dm³.
a dm³ is 1000 cm³
Hence we should have

1000cm³*cm*cm^-2*s^-2*g=kg*m²*s^-2

1000cm²*g=kg*m²

1000cm² are 0,1 m² so,

0,1m²*g=m²*kg

0,1g=kg

10^-1g= 10³*g

divided thorugh 10^-1

g=10⁴*g <-------------- O_O

but now?

[Borek]

Pressure is - in general - force/surface.

Now you know units are OK. Don't bother with conversion factor - it is part of the numerical value of R.

[Yggdrasil]

I was using atmospheres (atm). But I remember that 1 atm is: dyne per cm²

1 atmosphere = 1,013,250 dynes per cm² = 101.3 kPa = 101.3x10³ N/m².

Using this conversion you should see that the two values of R are in fact equivalent.

[Valeder]

1 atmosphere = 1,013,250 dynes per cm² = 101.3 kPa = 101.3x10³ N/m².

Using this conversion you should see that the two values of R are in fact equivalent.

Of course, I had confused bar (which is indeed dyne*cm^-2) with atmosphere which isnùt a unit of measure of the cgs system but a pratical unit of measure.

So that:

101.300 Pa = 10⁵ Pa = 10⁵ N*m^-2

8,2*10^-2*10⁵*N*m^-2*10^-3*m³*mol^-1*K^-1= 8,3 N*m*mol^-1*K^-1

then at left we have 10^5-3-2=10⁰=1
semplifying moles and Kelvins

N*m^-2*m³=N*m
N*m=N*m

and everything comes happily 

Solution: I must learn to read better the books and in particular not to skip or read fastly the unit of measure section 

Thanks to everybody