Gas Density — Ideal Gas Method
Start from the density definition \( d = \dfrac{m}{V} \). For a gas that obeys the ideal gas
equation \( PV = nRT \) and using \( n = \dfrac{m}{M} \) (with \(M\) the molar mass), we can
connect density to the state variables:
\[
PV = nRT,\qquad n=\dfrac{m}{M}
\;\;\Longrightarrow\;\;
\boxed{\, d \;=\; \dfrac{m}{V} \;=\; \dfrac{M \cdot P}{R \cdot T} \,}
\]
Rearrangements (solve any variable)
\[
\begin{aligned}
d &= \dfrac{M \cdot P}{R \cdot T} \\
M &= \dfrac{d \cdot R \cdot T}{P} \\
P &= \dfrac{d \cdot R \cdot T}{M} \\
T &= \dfrac{M \cdot P}{d \cdot R}
\end{aligned}
\]
Units & constants (convert before calculating)
Use SI internally: \(P\,[\mathrm{Pa}],\; T\,[\mathrm{K}],\; M\,[\mathrm{kg\,mol^{-1}}],\; d\,[\mathrm{kg\,m^{-3}}]\).
The gas constant:
\[
R = 8.314462618\ \mathrm{Pa\cdot m^{3}\cdot mol^{-1}\cdot K^{-1}}
= 0.082057\ \mathrm{atm\cdot L\cdot mol^{-1}\cdot K^{-1}}
= 62.3637\ \mathrm{L\cdot Torr\cdot mol^{-1}\cdot K^{-1}}
\]
\[
1\ \mathrm{atm}=101325\ \mathrm{Pa},\quad
1\ \mathrm{bar}=10^{5}\ \mathrm{Pa},\quad
1\ \mathrm{mmHg}\approx 133.322\ \mathrm{Pa},\quad
1\ \mathrm{Torr}=133.322\ \mathrm{Pa},\quad
1\ \mathrm{psi}=6894.757\ \mathrm{Pa}
\]
\[
1\ \mathrm{L}=10^{-3}\ \mathrm{m^{3}},\qquad
1\ \mathrm{mL}=1\ \mathrm{cm^{3}}=10^{-6}\ \mathrm{m^{3}}
\]
\[
T\ (\mathrm{K}) \;=\; T\ (^\circ\mathrm{C}) + 273.15
\]
Trends & intuition
- Directly proportional to \(P\) and \(M\): higher pressure or larger molar mass \(\Rightarrow\) higher density.
- Inversely proportional to \(T\): hotter gas \(\Rightarrow\) lower density (at fixed \(P\)).
- Gas densities are typically reported in \(\mathrm{g\,L^{-1}}\) (very small compared with liquids/solids).
Worked example (O\(_2\) at 298 K and 0.987 bar)
Given \(M=32.00\ \mathrm{g\,mol^{-1}}\), \(T=298\ \mathrm{K}\), \(P=0.987\ \mathrm{bar}\).
\[
d \;=\; \dfrac{M \cdot P}{R \cdot T}
\;=\; \dfrac{32.00\ \mathrm{g\,mol^{-1}} \cdot 0.987\ \mathrm{bar}}
{0.08314\ \mathrm{bar\cdot L\cdot mol^{-1}\cdot K^{-1}} \cdot 298\ \mathrm{K}}
\;\approx\; 1.27\ \mathrm{g\,L^{-1}}
\]
STP shortcut
At STP, the molar volume is approximately \(22.7\ \mathrm{L\,mol^{-1}}\). Thus
\[
d_{\text{STP}} \;\approx\; \dfrac{M}{22.7}\ \mathrm{g\,L^{-1}}.
\]
Corrections & caveats
- Real gases: include a compressibility factor \(Z\):
\[
PV = n Z R T \quad\Rightarrow\quad d=\dfrac{M \cdot P}{Z \cdot R \cdot T}.
\]
Deviations from \(Z=1\) grow at high \(P\) and low \(T\).
- Dry gas pressure: if the gas is moist, use \(P_{\text{dry}} = P_{\text{total}} - P_{\mathrm{H_2O}}\).
- Absolute pressure: convert gauge readings to absolute before use.
Quality checks
- Units reduce to density units (e.g., \(\mathrm{g\,L^{-1}}\) or \(\mathrm{kg\,m^{-3}}\)).
- Magnitude sanity: at room \(T\) and near 1 atm, many gases are \(1\)–\(2\ \mathrm{g\,L^{-1}}\) (air \(\approx\) \(1.2\ \mathrm{g\,L^{-1}}\)).
- Round only at the end; match the least certain input (your calculator lets you choose 2–8 s.f.).
Common pitfalls
- Using \(R\) with incompatible units (e.g., mixing Pa with \(R=0.082057\)).
- Forgetting to convert \(^\circ\mathrm{C}\) to \(\mathrm{K}\).
- Accidentally using gauge pressure or neglecting water vapor pressure.
