Manometers, barometers: Question bank — Physics deep dive

Definition Symbols used on this page (learn these first)

$ρ$ (Greek "rho") = density of the fluid, meaning mass packed per unit volume (kg per cubic metre). Heavy fluids like mercury have large $ρ$ ; see Density and Specific Gravity.
$g$ = gravitational acceleration, how hard gravity pulls each kilogram down ( $\approx 9.8 m/s^{2}$ on Earth). It sets how much a column of fluid weighs.
$h$ = the vertical height of a fluid column or the vertical gap between two surfaces (metres). Only vertical counts — slanted length does not.
$P_{a t m}$ = atmospheric (air) pressure pushing on any open surface ( $\approx 1.013 \times 1 0^{5} Pa$ ); see Atmospheric Pressure.
$P_{g a s}$ = the pressure of the trapped gas or vessel that a manometer is measuring.
$P_{ab s}$ = absolute pressure (measured from a true vacuum); $P_{g a ug e} = P_{ab s} - P_{a t m}$ = how much you're above atmosphere.

Before you start, keep three anchors in mind:

Only vertical height counts — the fluid's sideways travel is free.
Same connected fluid, same level ⇒ same pressure (from Pascal's Law).
Barometer top is vacuum (absolute); open manometer's arm meets air (gauge).

True or false — justify

A wider barometer tube gives a taller mercury column.

False.

h = P_{a t m} / (ρ g)

has no area in it; the extra mercury a fat tube holds also sits on more area, so the height is identical.

Tilting a barometer tube (top still sealed) changes the reading in mmHg.

False for the vertical height (see the staircase figure s02) — the mercury climbs along the slanted tube but its vertical column stays

0.76 m

; only the slanted length grows. The pressure it balances is unchanged.

In a U-tube open manometer, the two liquid surfaces are always at different heights.

False. If

P_{g a s} = P_{a t m}

then

ρ g h = 0

, so

h = 0

and both surfaces sit level — the middle snapshot of figure s03.

A barometer reads gauge pressure.

False. Its sealed top is a near-vacuum (

P \approx 0

), so

P_{a t m} = ρ g h

is the absolute atmospheric pressure — nothing was subtracted off.

Doubling

g

(say on a heavier planet) with the same atmosphere doubles the barometer's mercury height.

False.

h = P_{a t m} / (ρ g)

: more

g

makes each mm of mercury heavier, so the column gets shorter, not taller —

h

scales as

1/ g

Two points at the same depth in the same still liquid have the same pressure even if one is under a narrow neck and one under a wide bulb.

True. Same fluid, same vertical level ⇒ same pressure; shape and width are irrelevant (the hydrostatic paradox).

Replacing mercury with water in an open manometer changes what "gauge" means.

False. It still reads gauge pressure

P_{ab s} - P_{a t m}

; only the height

h

needed to show that gauge pressure changes (water gives a much taller, more sensitive column).

If the gas arm liquid is higher than the open arm, the gas is below atmospheric pressure.

True — the right snapshot of figure s03. The open air pushed its own side down and the gas side up, so

P_{g a s} = P_{a t m} - ρ g h < P_{a t m}

In an accelerating (non-inertial) frame, a manometer still balances with

g

alone.

False. In a frame with linear acceleration

a

the fluid feels an effective gravity

g_{eff} = g^{2} + a^{2}

tilted from vertical; surfaces align perpendicular to

g_{eff}

, so you must use

g_{eff}

and its direction, not plain vertical

g

Spot the error

"The pressure at the bottom of a tank depends on how much water is in it, so a bigger tank means higher bottom pressure."

Error: pressure depends on depth

h

, not total volume. Two tanks filled to the same depth have equal bottom pressure regardless of width or total water (same reasoning as the barometer figure s01).

"For an inclined manometer with liquid length

L = 0.20 m

3 0^{\circ}

, use

h = 0.20 m

ρ g h

Error: only the vertical rise pushes against gravity (staircase figure s02). Use

h = L sin θ = 0.20 \times 0.5 = 0.10 m

"A barometer and an open manometer both use

ρ g h

, so both give absolute pressure."

Error: the top of each column differs (compare the two halves of figure s01). Barometer top is vacuum ⇒ absolute; open manometer's other arm sits at

P_{a t m}

⇒ its

ρ g h

is only the gauge (difference) pressure.

"To read the gas pressure, add up the height of liquid in both arms of the U-tube."

Error: what matters is the difference between the two free surfaces (the gap

h

in figure s03), not the sum. The shared lower fluid cancels out when you equate pressures at the reference level.

"Because pressure pushes in all directions, the horizontal pressure differences also lift the manometer fluid."

Error: only the vertical pressure change

ρ g h

does mechanical work against gravity; along a horizontal level the pressure is equal on both sides and cancels (green segments in figure s02).

"A vacuum has negative pressure, so the top of a barometer pulls the mercury up."

Error: vacuum pressure is

\approx 0

, not negative. The mercury is pushed up from below by

P_{a t m}

on the open dish (see the upward arrow in figure s01), not pulled from above.

Why questions

Why does the cross-sectional area

A

cancel in the hydrostatic derivation?

Because pressure is force per area: the extra weight of fluid (

\propto A

) sits on an equally larger area (

\propto A

), so their ratio — the pressure — is area-independent.

Why is mercury preferred over water in a barometer?

Mercury is

13.6 \times

denser, so from

h = P / ρ g

the column is

13.6 \times

shorter —

0.76 m

instead of

\sim 10.3 m

, which fits on a bench (see Density and Specific Gravity).

Why must the two barometer/dish points be at the same horizontal level to equate their pressures?

In a connected still fluid, a pressure difference at equal height would drive sideways flow; equilibrium forbids flow, so equal level ⇒ equal pressure (Pascal's Law).

Why does an open-tube manometer read gauge and not absolute pressure?

One arm is exposed to

P_{a t m}

, so the height difference measures how much the gas exceeds or falls short of atmospheric — that difference is the gauge pressure by definition.

Why does only vertical height, not path length, appear in

ρ g h

Because pressure change is the work gravity does per unit volume as fluid moves; horizontal moves are at constant gravitational potential (no work), so only the vertical drop counts — exactly the green-vs-red split in the staircase figure s02.

Why can't you use

ρ g h

directly for a fast-flowing fluid in the tube?

ρ g h

assumes static equilibrium (no motion). Moving fluid has kinetic-energy terms handled by Bernoulli's Equation instead.

Why does the barometer measure atmospheric pressure but the diver's ear pain measures water pressure — using the same law?

Both are

ρ g h

from the same free surface; the barometer's "depth" is a mercury column balancing air, the diver's is water depth — same physics of weight-per-area from Atmospheric Pressure.

Why does an accelerating manometer's fluid surface tilt?

Because in the fluid's frame the effective gravity is the vector sum of real gravity (down) and the inertial "pull" (

- a

), so the free surface sets itself perpendicular to that tilted total force — the surface is always flat relative to effective gravity, not to the ground.

Edge cases

What does an open manometer read when

P_{g a s} = P_{a t m}

exactly?

h = 0

: both surfaces are level (middle of figure s03), so gauge pressure is zero and the device shows no height difference.

What happens to the mercury if a barometer's sealed top somehow leaks and lets air in?

The trapped air adds pressure on top, so the column drops; it no longer reads

P_{a t m}

because the top is no longer a vacuum.

If a manometer is used with a gas at zero absolute pressure (a vacuum vessel), what is

h

P_{g a s} = 0

gives gauge

= - P_{a t m}

, so the open arm falls by

h = P_{a t m} / (ρ g)

— the maximum the atmosphere can depress that side. It behaves like a barometer.

Can a water manometer measure a pressure difference of

1 atm

in a normal-sized lab tube?

Not practically — it would need

\sim 10.3 m

of vertical water. That's exactly why large pressures use dense fluids like mercury.

Two connected arms hold different liquids (oil over water). Is same-level-same-pressure still true across the boundary?

Only within one continuous fluid. Across an interface you must account for each fluid's own

ρ g h

; equal pressure holds at the shared interface level, not blindly across different densities.

At the exact liquid surface open to air, what is the gauge pressure?

Zero. Gauge is measured relative to

P_{a t m}

, and the free surface sits at

P_{a t m}

, so

P_{g a ug e} = ρ g (0) = 0

On a mountain top where

P_{a t m}

is lower, does the same manometer give a larger or smaller

h

for the same gas?

Larger gauge height if the gas stays fixed:

h = (P_{g a s} - P_{a t m}) / (ρ g)

grows as

P_{a t m}

drops — the gas now exceeds the thinner air by more.

A sealed barometer sits in a lift accelerating upward at

a

. Does the mercury column rise or fall?

It rises. Effective gravity becomes

g + a

, so each mm of mercury weighs more; but the air it balances is also heavier — for a barometer measuring the same

P_{a t m}

h = P_{a t m} / (ρ (g + a))

actually shortens. (Watch which quantity is held fixed.)

A U-tube manometer in free fall (

g_{eff} = 0

) — what does it read?

Nothing meaningful: with no effective gravity there is no

ρ g h

term, so any height difference no longer corresponds to a pressure — the device stops working.

Connections — how each linked idea plugs into these traps

Parent: Manometers & Barometers — the full derivations these traps test.