Question

In working high-pressure geochemistry problems it is usually more

convenient to express volumes in units of kJ/kbar. Work out the conversion factor

between this unit and m³

Short answer

The conversion factor is$\frac{1\mathrm{kJ}}{\mathrm{k}·\mathrm{bar}}={10}^{-5}{\mathrm{m}}^3$

Step-by-step solution

Step 1. Given information 

We used kJ/kbar as the unit of pressure in the high-pressure geochemical problem.

$$\begin{gathered} \frac{1\mathrm{kJ}}{\mathrm{k}\text{bar}}=1·\frac{{10}^3·\mathrm{J}}{{10}^3·\mathrm{bar}} \\ =1·\frac{\mathrm{J}}{\text{bar}} \end{gathered}$$

Calculation

Using the conversion between the units of pressure 1 bar = 10⁵ Pa, we get

$$\begin{gathered} =\frac{1\mathrm{J}}{\mathrm{bar}}\times \frac{\mathrm{lbar}}{{10}^5\mathrm{N}/{\mathrm{m}}^2} \\ ={10}^{-5}\frac{\mathrm{J}}{\mathrm{N}/{\mathrm{m}}^2} \\ ={10}^{-5}\times \frac{\mathrm{N}·\mathrm{M}}{\mathrm{N}/{\mathrm{m}}^2} \\ ={10}^{-5}{\mathrm{m}}^3 \\ \therefore \frac{1\mathrm{kJ}}{\mathrm{k}·\mathrm{bar}}={10}^{-5}{\mathrm{m}}^3 \end{gathered}$$