I have not done any physics in ages and have recently started studying it. The first chapter in my book deals with the ideal gas constant:


It is rewritten as:


When I write it in SI units, it looks like this:

$$8.31\frac{\rm J}{\rm mol\,K} = \frac{(\rm Pa)(m^3)}{\rm (mol)(K)}$$

How does $\rm (Pa)(m^3)$ translate into joules?

  • $\begingroup$ Note there are two forms of the ideal gas constant - the universal constant and the specific constant. For the equation and units you have expressed the $R$ is the universal gas constant, so it relates energy to temperature for any ideal substance in terms of mols. The answers you see below are all good and fine mathematically and unit-wise, but the concise meaning of $R$ is that it relates the proportionality of energy to temperature for an amount of gas particles. $\endgroup$
    – docscience
    Jan 9, 2015 at 15:55

2 Answers 2


The joule is the amount of energy needed to apply one newton of force for a distance of one meter: $$ \rm J=N\cdot m=\frac{kg\,m^2}{s^2}\tag{1} $$ Where the 2nd equality comes from the definition of the newton (mass times acceleration): $\rm N=kg\,m/s^2$. The pascal is defined as one newton of force applied to a one-square-meter area: $$ \rm Pa=\frac{N}{m^2}=\frac{kg}{m\,s^2}\tag{2} $$ Comparing (1) and (2) (specifically the last two equalities), we see that $$ \rm Pa\cdot m^3=\frac{kg}{m\,s^2}\cdot m^3=\frac{kg\,m^2}{s^2}\equiv J $$


If a constant pressure of $1\,\rm Pa$ is exerted on a piston, and pushes it back so as to liberate a volume of $1\,\rm m^3$, then the work done by pressure on the piston amounts to $1\,\rm J$.


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