### Why is a Liter-Atmosphere a Unit of Energy?

Isn't That What a Joule is?

Back to PV Work

Return to Thermochemistry menu

That's right and they are equal to each other. The proof depends on using all the proper SI definitions. From the study of gases, we know this to be true:

PV = nRT

Rearranging:

R = (PV) / (nT)

Depending on the units chosen R can have different values. Two of the most popular are **L-atm / mol-K** and **J / mol-K.** So what I propose to do is an analysis of the units in the numerator and change them over from liter-atmospheres to Joules.

The first thing is to use Pascals as the pressure unit rather than atmospheres. Those of you who don't think I can do this, please consider that 101,325 Pa = 1 atm. When I substitute Pa for atm., all I change is the numerical value of R. I also substituted m^{3} for the volume, remembering that 1 L = 1 dm^{3}. That means the units on the numerator are:

(m^{3}) (Pa)

Now, Pascals are a pressure unit and keep in mind that pressure equals force per unit area. Since a Pascal equals a Newton per square meter, we have this now:

(N / m^{2}) (m^{3})

That Newton per square meter stuff is coming to you out of the clear blue sky, I know that. You can study up on it later. For the moment, trust your friendly ChemTeam!!

Next, we need the definition of a Newton. It is the SI unit for force and it is:

the amount of net force that gives an acceleration of one meter per second squared

to a body with a mass of one kilogram.

and the units would be:

(kg m) / s^{2}

I'll put the units for a Newton in place of the symbol N to get:

[(kg m) / (s^{2} m^{2}] (m^{3})

The m and m^{3} are both in the numerator to give m^{4} and the m^{2} cancels with it to give m^{2} (meter squared) in the numerator, learving this as a final answer:

(kg m^{2}) / s^{2}

which is the unit (from mv^{2}) known to be the unit for Joules.

Q.E.D.

Back to PV Work

Return to Thermochemistry menu