What are the units of P and V when R is expressed in terms of J/(mol K)?

I have learnt the ideal gas equation $PV=nRT$. However, I have some doubts regarding the units of the quantities in this equation. 


When we take the gas constant $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$, what should be the units of $P$ and $V$? 




Would you agree that pressure is expressed as a force acting on a surface? If so, let's juggle with units and start with the *pascal*. 
$1\ \mathrm{Pa} = 1\ \mathrm{Nm^{-2}}$
will certainly be familiar to you ;)


Would you further agree that we can express the *joule* as $1\ \mathrm J = 1\ \mathrm{Nm}$?


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The units for pressure $P$ are pascals ($\mathrm{Pa}$) and volume $V$ is metres cubed ($\mathrm m^3$). The proof can be found by the derivation provided from the University of Waterloo's page [The Ideal Gas Law](http://www.science.uwaterloo.ca/~cchieh/cact/c120/idealgas.html), reproduced below:


Rearrange the formula to make $R$ the subject, hence:


$$R = \frac{PV}{nT}$$


Now rewrite the Ideal Gas Law in terms of its units:


$$\mathrm{J\ mol^{-1}\ K^{-1}} = \frac{\mathrm{Pa\ m^3}}{\mathrm{mol\ K}}$$


or


$$\mathrm{J\ mol^{-1}\ K^{-1}} = \mathrm{Pa\ m^3\ mol^{-1}\ K^{-1}}$$


Given that $\mathrm{Pa} = \mathrm{N\ m^{-2}}$, rewrite that into the equation:


$$\mathrm{J\ mol^{-1}\ K^{-1}} = \mathrm{N\ m^{-2}\ m^3\ mol^{-1}\ K^{-1}}$$


Now the term $\mathrm{N\ m^{-2}\ m^3}$ can be simplified to $\mathrm{N\ m} = \mathrm J$ as work (unit = $\mathrm J$) = force x distance ($\mathrm{N\ m}$)


Hence, the right hand side becomes $\mathrm{J\ mol^{-1}\ K^{-1}}$ as required


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