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I am trying to calculate the molar volume of a perfect gas under standard ambient temperature and pressure, but I'm having problems with the units.

$$\begin{align} V_m = \dfrac{RT}{p} &= \dfrac{(8.314 \ \text{J K$^{-1}$ mol$^{-1}$})(298.15 \ \text{K})}{10^5 \text{Pa}} \\ &= 0.024789 \ \text{N$^{-1}$ m$^2$ mol$^{-1}$ Nm} \\ &= 0.024789 \ \text{m$^{3}$ mol$^{-1}$} \\ &= 0.24789 \ \text{dm$^{3}$ mol$^{-1}$} \ \ \ (\text{1 metre $=$ 10 decimetres)} \end{align}$$

I am told that it should be $24.789 \ \text{dm$^3$ mol$^{-1}$}$.

I would greatly appreciate it if someone would please take the time to help me understand what I'm doing incorrectly.

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  • $\begingroup$ 1 m^N = 10^N dm^N, so 1 m = 10 dm, 1m^2 = 100 dm^2, 1 m^3 = 1000 dm^3 $\endgroup$
    – Poutnik
    Commented Jun 14, 2020 at 14:10

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$$\begin{align} V_m = \dfrac{RT}{p} &= \dfrac{(8.314 \ \text{J K$^{-1}$ mol$^{-1}$})(298.15 \ \text{K})}{10^5 \text{Pa}} \\ &= 0.024789 \ \text{N$^{-1}$ m$^2$ mol$^{-1}$ Nm} \\ &= 0.024789 \ \text{m$^{3}$ mol$^{-1}$} \end{align}$$

Here's where you're going wrong. $1\ m = 10\ dm$. So, $$1\ m^3 = 1000\ dm^3\ !!!$$

That automatically makes it $24.789 \ \text{dm$^3$ mol$^{-1}$}$

Cheers!

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Your calculation is correct, except the last line. One meter is $10$ decimeter. But one cubic meter is $1000$ cubic decimeter. So $0.0248$ m$^3$ = $24.8$ dm$^3$. That's all.

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