Barometric Pressure — Theory
A barometer measures the absolute pressure of the atmosphere. In a classic (Torricelli)
mercury barometer, a long tube filled with liquid mercury is inverted into a mercury reservoir.
The space at the top of the tube is (nearly) a vacuum, so the only force supporting the column
is the atmosphere pressing on the reservoir. At equilibrium the hydrostatic head balances the atmosphere:
\[
P_{\text{atm}} \;=\; \rho \cdot g \cdot h
\]
- \(\rho\): liquid density (kg·m\(^{-3}\)) — for Hg, \(\rho_{\mathrm{Hg}}\approx 13.5951\ \text{g·cm}^{-3}\) at \(0^\circ\)C.
- \(g\): local gravitational acceleration (\(\text{m·s}^{-2}\)); standard \(9.80665\ \text{m·s}^{-2}\).
- \(h\): liquid column height (m).
Closed-end vs open-end tubes
- Closed-end (barometer): Vacuum above the column ⇒ equation above gives absolute atmospheric pressure.
- Open-end manometer: Gas pressure is compared to atmosphere; it typically gives a gauge difference, not absolute.
mmHg vs Torr
Two “millimeter of mercury” style units are widely seen:
- Torr — defined exactly as \(1\ \text{Torr} = \dfrac{1}{760}\ \text{atm}\).
- mmHg — the conventional pressure of a 1-mm Hg column computed from \(\rho g h\).
Using standard values gives \(1\ \text{mmHg} \approx 133.322\ \text{Pa}\).
Because \(\rho\) and \(g\) depend slightly on temperature and location, mmHg is not an exact fraction of atm.
In most chemistry contexts, Torr and mmHg are used interchangeably to 4–5 sig figs.
Exact and conventional equivalences
| Statement | Notes |
|---|
| \(1\ \text{atm} = 101\,325\ \text{Pa}\) | Exact (SI definition). |
| \(1\ \text{atm} = 760\ \text{Torr}\) | Exact (definition of Torr). |
| \(1\ \text{Torr} = \dfrac{101\,325}{760}\ \text{Pa}\) | Exact ⇒ \(133.322\,368...\ \text{Pa}\). |
| \(1\ \text{mmHg} \approx 133.322\ \text{Pa}\) | Conventional (via \(\rho g\) of Hg). |
Worked example (deriving 1 atm from a 760-mm Hg column)
Using standard \(\rho_{\mathrm{Hg}} = 13.5951\ \text{g·cm}^{-3}\), \(g = 9.80665\ \text{m·s}^{-2}\), \(h = 760.000\ \text{mm}\):
\[
\begin{aligned}
\rho_{\mathrm{Hg}} &= 13.5951\ \mathrm{g\,cm^{-3}} \;=\; 1.35951\times 10^{4}\ \mathrm{kg\,m^{-3}} \\
h &= 760.000\ \mathrm{mm} \;=\; 0.760000\ \mathrm{m} \\
g &= 9.80665\ \mathrm{m\,s^{-2}} \\
P_{\text{atm}} &= \rho g h \\
&= (1.35951\times 10^{4})\,(9.80665)\,(0.760000)\ \mathrm{kg\,m^{-1}\,s^{-2}} \\
&= 1.01325\times 10^{5}\ \mathrm{Pa} \\
&= 101.325\ \mathrm{kPa} \;\approx\; 1\ \mathrm{atm}
\end{aligned}
\]
How tall would a water barometer be?
Replace mercury with water (\(\rho \approx 1000\ \text{kg·m}^{-3}\)) and solve
\(h = \dfrac{P}{\rho g}\) at \(P = 1\ \text{atm}\):
\[
\begin{aligned}
h &= \frac{101\,325}{(1000)(9.80665)}\ \mathrm{m} \\
&\approx 10.33\ \mathrm{m}
\end{aligned}
\]
This is why practical barometers use dense mercury: the column is only about 760 mm, not 10.3 m.
Common pitfalls
- Confusing absolute pressure (barometer) with gauge pressure (manometer).
- Mixing Torr (exact) with mmHg (conventional) in high-precision work.
- For \(\rho g h\) computations, always convert: \(\rho\) to kg·m\(^{-3}\), \(h\) to m, \(g\) to m·s\(^{-2}\) to get Pa.
- Be consistent with significant figures; standards in this course use 3–4 sig figs unless the problem states otherwise.
Quick reference
| Unit | Pa |
|---|
| 1 atm | 101 325 (exact) |
| 1 Torr | 101 325/760 (exact) |
| 1 mmHg | ≈ 133.322 |
| 1 bar | 100 000 |
| 1 psi | ≈ 6 894.757 |
| Liquid | \(\rho\) (kg·m\(^{-3}\)) | Height at 1 atm |
|---|
| Hg | 13 595 | ≈ 0.760 m |
| Water | 1 000 | ≈ 10.33 m |
| Seawater | 1 025 | ≈ 10.08 m |
Temperature and local \(g\) slightly change \(\rho g h\); the calculator lets you adjust both for more accurate barometric predictions.
For conversions, the tool uses the exact SI definitions (atm, Torr) and the conventional factor for mmHg.
