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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 $V$?

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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, 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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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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