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Given the value 380 PaK^-1, how do I convert Pa to Jdm-3?

I know that 1 joule = 1 Nm, and that 1 pascal = 1 N/m^2.

But where does the dm^-3 (or L) come into this? I don't know where to start.

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  • $\begingroup$ The conversion factor is 1. $\endgroup$ – Chet Miller Oct 29 '15 at 0:50
  • $\begingroup$ Thank you, but where can I find workings to justify this? $\endgroup$ – Hvb123 Oct 29 '15 at 0:51
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Derived units are defined as products of powers of the base units. Some of the coherent derived units in the SI are given special names and symbols.

The joule (unit symbol: $\mathrm J$) and the pascal (unit symbol: $\mathrm{Pa}$) are such coherent derived units.

Expressed in terms of SI base units, one joule is

$$\mathrm{1\ J=1\ \frac{kg\ m^2}{s^2}}$$

and one pascal is

$$\mathrm{1\ Pa=1\ \frac{kg}{m\ s^2}}$$

(see: BIPM: The International System of Units (SI))

Hence,

$$\mathrm{1\ \frac J{m^3}=1\ \frac{\frac{kg\ m^2}{s^2}}{m^3}=1\ \frac{kg\ m^2}{m^3\ s^2}=1\ \frac{kg}{m\ s^2}=1\ Pa}$$

SI prefixes may be used in forming the decimal multiples and submultiples of units. Note that, when SI prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor.

$$\begin{align} \mathrm{1\ m}&=\mathrm{10\ dm}\\[6pt] \mathrm{1\ m^3}&=\mathrm{1000\ dm^3}\\[6pt] \mathrm{\frac1{m^3}}&=\mathrm{\frac1{1000\ dm^3}}=\mathrm{0.001\ \frac1{dm^3}}\\[6pt] \mathrm{1\ \frac J{m^3}}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt] \mathrm{1\ Pa}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt] \mathrm{1000\ Pa}&=\mathrm{1\ \frac J{dm^3}} \end{align}$$

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