Liquid Pressure — Chemistry Context
In fluids at rest, pressure is the normal force per unit area. In chemistry we encounter pressure as
hydrostatic pressure from a liquid column (manometers, barometers, density columns) and as
mechanical pressure from pistons or syringes.
Key relations
\[
P=\frac{F}{A}
\qquad\text{and}\qquad
P=\rho g h
\]
where \(P\) is pressure, \(F\) force, \(A\) area, \(\rho\) density, \(g\) acceleration of gravity, \(h\) liquid height.
Deriving \(P=\rho g h\)
Consider a vertical cylinder of cross-section \(A\) filled with a liquid of density \(\rho\) to a height \(h\).
The hydrostatic (gauge) pressure at the bottom equals the weight of the liquid above divided by its area:
\[
P=\frac{W}{A}=\frac{mg}{A} = \frac{(\rho V)g}{A}
=\frac{\rho (Ah) g}{A} = \rho g h.
\]
This result depends only on density and height (not on vessel shape). It gives gauge pressure, i.e., the
pressure above ambient (atmospheric).
Gauge vs Absolute Pressure
\[
P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}},\qquad
P_{\text{atm}} \approx 1\ \text{atm} = 101{,}325\ \text{Pa}.
\]
Manometer readings from a liquid column (e.g., water, mercury) are typically gauge. To convert to absolute,
add \(P_{\text{atm}}\) if the instrument is referenced to the atmosphere.
Common Units & Conversions
| Pressure |
|---|
| 1 atm | = 101 325 Pa = 101.325 kPa ≈ 1.01325 bar |
| 1 bar | = 100 000 Pa ≈ 0.986923 atm |
| 1 mmHg (torr) | ≈ 133.322 Pa |
| 1 psi | ≈ 6 894.757 Pa |
| 1 cmH2O | ≈ 98.0665 Pa |
| 1 mmH2O | ≈ 9.80665 Pa |
| Density |
|---|
| 1 g·mL−1 | = 1000 kg·m−3 |
| 1 g·cm−3 | = 1000 kg·m−3 |
| 1 kg·L−1 | = 1000 kg·m−3 |
Typical Liquid Densities (25 °C)
| Liquid | ρ (g·mL−1) | ρ (kg·m−3) |
|---|
| Water | 1.000 | 1000 |
| Seawater | ≈ 1.025 | ≈ 1025 |
| Ethanol | ≈ 0.789 | ≈ 789 |
| Glycerol | ≈ 1.261 | ≈ 1261 |
| Mercury | ≈ 13.534 | ≈ 13 534 |
Worked Examples
1) Pressure from a Water Column
Find the gauge pressure at a depth of \(h=0.10\ \text{m}\) in water (\(\rho = 1.000\ \text{g·mL}^{-1}\)).
\[
\rho = 1000\ \mathrm{kg\,m^{-3}},\quad g=9.80665\ \mathrm{m\,s^{-2}}
\quad\Rightarrow\quad
P = \rho g h = (1000)(9.80665)(0.10) \approx 981\ \mathrm{Pa}
= 0.981\ \mathrm{kPa}.
\]
Absolute pressure at that depth would be \(P_{\text{abs}} \approx 101.325\ \text{kPa} + 0.981\ \text{kPa}\).
2) Column Height for a Given Pressure (Manometer Idea)
What height of mercury (\(\rho \approx 13{,}534\ \text{kg·m}^{-3}\)) corresponds to 1 atm (gauge)?
\[
h = \frac{P}{\rho g} = \frac{101{,}325}{(13{,}534)(9.80665)} \approx 0.760\ \text{m} = 760\ \text{mm}.
\]
3) Mechanical Pressure from a Piston
A 125 N plunger acts on a circular area \(A=3.0\ \text{cm}^2\) to pressurize a solution. What pressure is applied?
\[
A = 3.0\times10^{-4}\ \mathrm{m^2},\quad
P=\frac{F}{A}=\frac{125}{3.0\times10^{-4}}
\approx 4.17\times10^{5}\ \mathrm{Pa}=0.417\ \mathrm{MPa}\approx 4.12\ \mathrm{bar}.
\]
Tips & Pitfalls
- Gauge vs absolute: manometers and \( \rho g h \) give gauge pressure; add 1 atm for absolute if needed.
- Units: always convert heights to meters and densities to kg·m−3 for \( \rho g h\) in Pa.
- Liquid identity matters: using water vs mercury changes pressure for the same height by the density ratio.
- Chemistry use-cases: osmotic setups, gas collection over water (barometric correction), and titration burets (hydrostatic head) all rely on these relations.
