# Calculating the molar volume of a perfect gas under standard ambient temperature and pressure [duplicate]

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.

• 1 m^N = 10^N dm^N, so 1 m = 10 dm, 1m^2 = 100 dm^2, 1 m^3 = 1000 dm^3 Commented Jun 14, 2020 at 14:10

\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 $$24.789 \ \text{dm^3 mol^{-1}}$$
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.