Gauge vs Absolute Pressure: The Difference

Absolute pressure is the total push at a point, the atmosphere included. Gauge pressure is only the amount by which that total exceeds atmospheric pressure, which is what a tire gauge and the depth formula give you. At sea level the two differ by one atmosphere, about 100,000 pascals.

AP Physics: Unit 8 (topics 8.2 Pressure). Everything on this page comes from AP Physics 1 Topic 8.2, Pressure, in Unit 8, Fluids. Essential knowledge 8.2.A.1 defines pressure as the magnitude of the perpendicular force component exerted per unit area over a given surface area; 8.2.A.2 states that pressure is a scalar quantity; and 8.2.A.3 states that the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid. Under the second learning objective, 8.2.B.1 attributes fluid pressure to the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact; 8.2.B.2 states that the absolute pressure of a fluid at a given point is equal to the sum of a reference pressure P nought, such as the atmospheric pressure P atm, and the gauge pressure, with the relevant equation P equals P nought plus rho g h; and 8.2.B.3 gives the gauge pressure of a vertical column of fluid as rho g h. Both equations are printed in the Mechanics and Fluids panel of the Table of Information on the AP Physics 1 and the AP Physics 2 sheets alike, as two of the seven fluid lines at the foot of that panel's right-hand column. The constants box on both sheets prints 1 atm as 1.0 times 10 to the fifth newtons per square meter, equal to 1.0 times 10 to the fifth pascals. The ideal gas law PV equals nRT is printed only on the AP Physics 2 sheet, in its Thermal Physics panel, and its pressure is an absolute pressure. Topic 8.2 carries no boundary statement; Unit 8's only boundary statement is printed under Topic 8.4 and reads that all fluids will be assumed to be ideal, and all pipes are assumed to be completely filled by the fluid, unless otherwise stated. Unit 8 is weighted at 10 to 15 percent of the multiple-choice section across a suggested 12 to 17 class periods, and the suggested skills listed for Topic 8.2 are 1.C, 2.B, 2.C and 3.C.

The distinction, stated once

Both are pressures, both are scalars, both are measured in pascals. They differ in where zero sits.

Absolute pressure counts from a perfect vacuum. It is the whole push the fluid exerts at that point, and it is what any equation treating pressure as a physical state of the fluid requires.

Gauge pressure counts from the local atmosphere. It is the excess over what the surroundings already supply, and it is what almost every instrument you have handled actually reports.

Essential knowledge 8.2.B.2 states the relation in words: the absolute pressure of a fluid at a given point is equal to the sum of a reference pressure $P_0$ , such as the atmospheric pressure $P_{\text{atm}}$ , and the gauge pressure $P_{\text{gauge}}$ . Its relevant equation is the one on the sheet,

P = P_0 + \rho g h

and 8.2.B.3 gives the second half by itself: the gauge pressure of a vertical column of fluid is described by

P_{\text{gauge}} = \rho g h

Put them side by side and the arithmetic is trivial, $P = P_0 + P_{\text{gauge}}$ . All of the difficulty is in knowing which of the two a question has handed you and which one the next equation wants.

Read the wording of 8.2.B.2 once more, because one clause in it does real work. It says a reference pressure $P_0$ , such as the atmospheric pressure. $P_0$ is whatever pressure sits on top of your column, and it is atmospheric only when the surface is open to the air. Seal a tube and evacuate the space above the liquid and $P_0$ is zero, so gauge and absolute become the same number. That is why the printed equation carries $P_0$ and not $P_{\text{atm}}$ .

Gauge against absolute, row by row

Property	Gauge pressure	Absolute pressure
Symbol on the sheet	$P_{\text{gauge}}$	$P$
Where its zero sits	at the local atmospheric pressure	at a perfect vacuum
Printed equation	$P_{\text{gauge}} = \rho g h$ , from 8.2.B.3	$P = P_0 + \rho g h$ , from 8.2.B.2
What it measures	the excess over the surroundings	the total push at the point
Can it be negative	yes, a partial vacuum reads below zero	no
Scalar or vector	scalar, per 8.2.A.2	scalar, per 8.2.A.2
Unit	pascal, $1\ \text{Pa} = 1\ \text{N/m}^2$	pascal
At the open surface of a lake	zero	one atmosphere, $1.0 \times 10^5\ \text{Pa}$
What a tire gauge reads	yes	no
What the ideal gas law needs	no	yes
Depends on	depth and fluid density	depth, fluid density and the reference pressure

The row that decides most questions is the last but one. $PV = nRT$ is printed on the AP Physics 2 sheet with $P$ meaning the pressure of the gas, full stop, and a gas at rest in a vacuum chamber still has a pressure. Feeding it a number whose zero is set at one atmosphere is the same error as feeding a kelvin formula a Celsius temperature, and it fails for the same reason: the scale has the wrong origin. Celsius vs Kelvin is the same mistake in a different unit.

The row above it is the one that hides. A tire gauge, a blood pressure cuff, and the dial on a scuba tank all report the excess over the air around them, because each of them is really a spring or a diaphragm with the atmosphere pushing on its other face. Nothing on the instrument says so. A tire inflated to a gauge reading of $2.2 \times 10^5\ \text{Pa}$ holds air at an absolute $3.2 \times 10^5\ \text{Pa}$ , and a tire reading zero is not empty, it is at atmospheric pressure.

The one atmosphere between them, and where that number is printed

You do not have to remember the conversion, and you should not remember it from chemistry. The constants box at the top of the Table of Information prints it on every one of the four sheets you might sit with:

1\ \text{atm} = 1.0 \times 10^5\ \text{N/m}^2 = 1.0 \times 10^5\ \text{Pa}

Two significant figures, not the $101{,}325\ \text{Pa}$ you may have met elsewhere. Carrying the more precise figure is not wrong physics, but it produces answers whose last two digits the rest of your data cannot support, and it wastes time.

The AP Physics 1 constants box is short enough to quote in full. It holds four lines: the universal gravitational constant, the atmosphere line above, the magnitude of the acceleration due to gravity at Earth's surface as $g = 9.8\ \text{m/s}^2$ , and the magnitude of the gravitational field strength at Earth's surface as $g = 9.8\ \text{N/kg}$ . Atmospheric pressure is one of three distinct constants that whole course gets. The AP Physics 2 constants box is much longer and prints the identical atmosphere line.

Both pressure equations are printed on both algebra-based sheets, in the panel headed Mechanics and Fluids. The fluid equations are the last seven lines of that panel's right-hand column, and it is worth counting them so you know what you are and are not given:

$\rho = \dfrac{m}{V}$
$P = \dfrac{F_{\perp}}{A}$
$P = P_0 + \rho g h$
$P_{\text{gauge}} = \rho g h$
$F_b = \rho V g$
$A_1 v_1 = A_2 v_2$
$P_1 + \rho g y_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g y_2 + \frac{1}{2} \rho v_2^2$

Seven, and the third and fourth are the pair this page is about. You are handed both forms, which means the exam expects you to choose between them rather than to derive one from the other.

The case that separates them: a bubble rising from a lake bed

A bubble of gas sits on the bottom of a fresh-water lake, $20.0\ \text{m}$ down, with a volume of $1.00\ \text{cm}^3$ . It breaks free and rises slowly enough that its temperature never changes. What volume does it have at the surface?

The gas obeys $PV = nRT$ with $n$ and $T$ fixed, so $P_1 V_1 = P_2 V_2$ . Everything now turns on which pressures you write down.

Gauge pressures. At the bottom, $P_{\text{gauge}} = \rho g h = (1000)(9.8)(20.0) = 1.96 \times 10^5\ \text{Pa}$ . At the surface, the gauge pressure is zero, because gauge pressure is measured from the atmosphere and the surface is at the atmosphere. So $P_2 = 0$ , and $V_2 = P_1 V_1 / P_2$ divides by zero. The bubble expands without limit. That is not a rounding error or a near miss, it is a physically absurd answer, and it is the clearest possible signal that gauge pressure has no business in this equation.

Absolute pressures. At the bottom, $P_1 = P_0 + \rho g h = 1.0 \times 10^5 + 1.96 \times 10^5 = 2.96 \times 10^5\ \text{Pa}$ . At the surface, $P_2 = P_0 = 1.0 \times 10^5\ \text{Pa}$ . Then

V_2 = V_1 \frac{P_1}{P_2} = (1.00)\frac{2.96 \times 10^5}{1.0 \times 10^5} = 2.96\ \text{cm}^3

The bubble very nearly triples. That is the answer.

There is a third route, and it is the one students actually take, which is why it is worth writing out. Use the gauge pressure at the bottom and the atmospheric pressure at the top, mixing the two scales without noticing:

V_2 = (1.00)\frac{1.96 \times 10^5}{1.0 \times 10^5} = 1.96\ \text{cm}^3

That is $1.96$ against a correct $2.96$ , low by $33.8\ \%$ . It looks like a plausible answer, it has the right units and the right order of magnitude, and nothing about it announces itself as wrong. The first route at least crashed. This one just quietly loses the marks.

Notice what makes the difference visible here and invisible elsewhere. The bubble problem compares pressures at two depths as a ratio. Every other fluid result you meet in Unit 8 compares them as a difference, and a difference is where the atmosphere cancels.

When it costs a mark

Substituting a gauge reading into $PV = nRT$ . The gas law needs the absolute pressure of the gas, always. If a problem says a gauge reads a value, or gives you a depth and expects you to use $\rho g h$ , add $P_0$ before the gas law sees the number.
Adding the atmosphere twice. $P = P_0 + \rho g h$ is already the absolute pressure. Writing $P_{\text{abs}} = P_0 + (P_0 + \rho g h)$ because you computed the depth term with the wrong equation first is the second commonest slip on this pair.
Measuring $h$ from the bottom. In both printed equations $h$ is the depth of the point below the surface where $P_0$ acts. In a tank of water $2.0\ \text{m}$ deep, a point $0.5\ \text{m}$ above the floor has $h = 1.5\ \text{m}$ , not $0.5\ \text{m}$ .
Carrying $101{,}325\ \text{Pa}$ and then reporting five figures. The sheet gives $1.0 \times 10^5\ \text{Pa}$ , so your atmospheric term is good to two significant figures and your answer cannot be better.
Treating pressure as directional. 8.2.A.2 states flatly that pressure is a scalar quantity. It has no components and it does not point. What has a direction is the force $P A$ on a particular surface, which is perpendicular to that surface, per 8.2.A.1.
Reading the word pressure as a promise of which kind. Problems say gauge when they mean gauge and usually say nothing when they mean absolute, but a depth, a manometer height, or an instrument reading is a gauge quantity whatever the sentence calls it. Decide from the physical setup, not the noun.
Using absolute pressure in a net-force calculation where the atmosphere acts on both faces. It is not wrong, it just makes you subtract two large numbers to get a small one, which invites arithmetic errors. The gauge value gives the same net force directly.

When they coincide, and why that lulls you

Almost every result in AP Physics 1 Unit 8 depends on a pressure difference, and in a difference the reference pressure cancels. That is why a student can work through the whole fluids unit using gauge pressure and never be corrected.

Take the buoyant force, derived from pressures rather than quoted. A cube of side $0.20\ \text{m}$ floats fully submerged in water with its top face $1.00\ \text{m}$ below the surface, so its bottom face is at $1.20\ \text{m}$ . Each face has area $0.040\ \text{m}^2$ .

With absolute pressures: the top face feels $P_0 + \rho g(1.00) = 1.098 \times 10^5\ \text{Pa}$ , pushing down with $4392\ \text{N}$ ; the bottom face feels $P_0 + \rho g(1.20) = 1.1176 \times 10^5\ \text{Pa}$ , pushing up with $4470.4\ \text{N}$ . The net upward force is $78.4\ \text{N}$ .

With gauge pressures: $9800\ \text{Pa}$ down over the top giving $392\ \text{N}$ , and $11760\ \text{Pa}$ up over the bottom giving $470.4\ \text{N}$ . Net upward force, $78.4\ \text{N}$ . Identical, because the two $P_0 A$ terms were equal and opposite and cancelled. And $F_b = \rho V g = (1000)(0.20)^3(9.8) = 78.4\ \text{N}$ agrees with both.

The same cancellation runs through the rest of the unit:

Bernoulli's equation. Add $P_0$ to $P_1$ and $P_2$ and the equation is unchanged, so gauge throughout and absolute throughout both work. What fails is gauge on one side and absolute on the other.
Manometers and open tubes. The answer is a height difference, which is a pressure difference.
The continuity equation. No pressure appears in it at all.
Apparent weight of a submerged object. It is weight minus buoyant force, and the buoyant force already cancelled the atmosphere.

So the distinction sits dormant for an entire unit and then matters suddenly, in exactly one situation: when a pressure enters an equation on its own rather than as part of a difference. The ideal gas law is the case you will meet. Buoyant force vs weight works the float-or-sink side of the same unit, where the cancellation is total and you never think about the atmosphere at all.

What the CED asks of Topic 8.2

Pressure is Topic 8.2 of AP Physics 1, inside Unit 8, Fluids, which carries $10$ to $15\ \%$ of the multiple-choice section across a suggested $12$ to $17$ class periods. The topic has two learning objectives and six pieces of essential knowledge between them, three under each.

8.2.A, describe the pressure exerted on a surface by a given force.

8.2.A.1 defines pressure as the magnitude of the perpendicular force component exerted per unit area over a given surface area, $P = F_{\perp}/A$ .
8.2.A.2 says pressure is a scalar quantity.
8.2.A.3 says the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid.

8.2.B, describe the pressure exerted by a fluid.

8.2.B.1 says the pressure exerted by a fluid results from the entirety of the interactions between the fluid's constituent particles and the surface with which those particles interact.
8.2.B.2 gives the absolute pressure as the sum of a reference pressure and the gauge pressure, with $P = P_0 + \rho g h$ .
8.2.B.3 gives the gauge pressure of a vertical column of fluid as $P_{\text{gauge}} = \rho g h$ .

Topic 8.2 prints no boundary statement. Unit 8 carries exactly one, and it sits under Topic 8.4: all fluids will be assumed to be ideal, and all pipes are assumed to be completely filled by the fluid, unless otherwise stated. Ideal vs real fluid unpacks what that single sentence buys you.

The suggested skills listed against Topic 8.2 are 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.B, calculate or estimate an unknown quantity with units from known quantities; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence. Skill 1.C is a hint about how this topic is examined: a graph of pressure against depth is a straight line of slope $\rho g$ , and its intercept is the whole of this page. Gauge pressure passes through the origin; absolute pressure crosses the axis at $P_0$ . Same slope, same fluid, different line.

One of the unit's own essential questions is why we do not feel the miles of air above us pushing us down. The answer is this distinction: your body is at a gauge pressure near zero because the air is on every side of you at once, while the absolute pressure inside you is a full atmosphere. Topic 8.2 has the CED framing in full.

A diver at ten meters, and the force on a viewing port

A diver descends to a depth of $10.0\ \text{m}$ in fresh water of density $1000\ \text{kg/m}^3$ . Take $g = 9.8\ \text{m/s}^2$ and $P_0 = 1.0 \times 10^5\ \text{Pa}$ . (a) Find the gauge pressure at that depth. (b) Find the absolute pressure, and express it in atmospheres. (c) A circular viewing port of radius $0.10\ \text{m}$ is set into the wall of a submersible at that depth, with ordinary air at $1.0 \times 10^5\ \text{Pa}$ inside. Find the force the water exerts on the port, the force the inside air exerts on it, and the net force on it.

(a) The gauge pressure of a vertical column of fluid is 8.2.B.3's $P_{\text{gauge}} = \rho g h$ . Substituting, $P_{\text{gauge}} = (1000)(9.8)(10.0)$ .
$(1000)(9.8) = 9800$ , and $9800 \times 10.0 = 98{,}000\ \text{Pa}$ , which is $9.8 \times 10^4\ \text{Pa}$ .
(b) The absolute pressure adds the reference pressure, per 8.2.B.2: $P = P_0 + \rho g h = 1.0 \times 10^5 + 9.8 \times 10^4$ .
$1.00 \times 10^5 + 0.98 \times 10^5 = 1.98 \times 10^5\ \text{Pa}$ . Dividing by the sheet's $1.0 \times 10^5\ \text{Pa}$ per atmosphere gives $1.98\ \text{atm}$ , so ten meters of water is very nearly one extra atmosphere. That is a useful sanity number to carry.
(c) The port's area is $A = \pi r^2 = \pi (0.10)^2 = 0.031416\ \text{m}^2$ .
The water pushes inward with the full absolute pressure, because water does not know about the air on the other side: $F_{\text{water}} = P A = (1.98 \times 10^5)(0.031416) = 6220\ \text{N}$ .
The cabin air pushes outward with $F_{\text{air}} = P_0 A = (1.0 \times 10^5)(0.031416) = 3142\ \text{N}$ .
The net force is inward, $6220.4 - 3141.6 = 3078.8\ \text{N}$ , or $3.1 \times 10^3\ \text{N}$ to two significant figures.
Now check that net force the short way: it should equal the gauge pressure times the area, because the gauge pressure is by definition the excess over the atmosphere sitting on the far side. $P_{\text{gauge}} A = (98{,}000)(0.031416) = 3078.8\ \text{N}$ . It matches exactly.
That agreement is the whole lesson of part (c). The individual forces on the two faces needed absolute pressures. Their difference needed only the gauge pressure, and got there in one line instead of three.

(a) $P_{\text{gauge}} = 9.8 \times 10^4\ \text{Pa}$ . (b) $P = 1.98 \times 10^5\ \text{Pa}$ , which is $1.98\ \text{atm}$ . (c) The water pushes in with $6.2 \times 10^3\ \text{N}$ and the cabin air pushes out with $3.1 \times 10^3\ \text{N}$ , for a net inward force of $3.1 \times 10^3\ \text{N}$ , equal to $P_{\text{gauge}} A$ .

The rising bubble, worked three ways

A bubble of gas of volume $1.00\ \text{cm}^3$ sits at the bottom of a fresh-water lake $20.0\ \text{m}$ deep. It rises to the surface slowly enough that its temperature is unchanged and no gas enters or leaves it. Take $\rho = 1000\ \text{kg/m}^3$ , $g = 9.8\ \text{m/s}^2$ and $P_0 = 1.0 \times 10^5\ \text{Pa}$ . (a) Find the volume at the surface using absolute pressures. (b) Show what happens if you use gauge pressures throughout. (c) Find the answer a student gets by mixing the two, and state the percentage error.

The amount of gas and the temperature are both fixed, so $PV = nRT$ collapses to $P_1 V_1 = P_2 V_2$ between the two states. The only question is which pressures to write.
(a) At the lake bed, 8.2.B.2 gives the absolute pressure $P_1 = P_0 + \rho g h$ . The depth term is $(1000)(9.8)(20.0) = 196{,}000\ \text{Pa} = 1.96 \times 10^5\ \text{Pa}$ .
So $P_1 = 1.0 \times 10^5 + 1.96 \times 10^5 = 2.96 \times 10^5\ \text{Pa}$ . At the surface the fluid column above the bubble has zero height, so $P_2 = P_0 = 1.0 \times 10^5\ \text{Pa}$ .
Rearranging, $V_2 = V_1 \dfrac{P_1}{P_2} = (1.00)\dfrac{2.96 \times 10^5}{1.0 \times 10^5} = 2.96\ \text{cm}^3$ .
(b) Now the gauge route. At the bed, $P_{\text{gauge}} = 1.96 \times 10^5\ \text{Pa}$ as computed above. At the surface, the gauge pressure is $\rho g (0) = 0$ .
That gives $V_2 = (1.00)(1.96 \times 10^5)/0$ , which is undefined. Physically it says the bubble expands without bound, which is false, and the failure is not an approximation going bad. It is the equation reporting that the pressure scale has the wrong zero.
(c) The mixed route, which is the one that actually appears in student work: gauge at the bottom because that is what $\rho g h$ gave, and atmospheric at the top because the surface is obviously at one atmosphere.
$V_2 = (1.00)\dfrac{1.96 \times 10^5}{1.0 \times 10^5} = 1.96\ \text{cm}^3$ .
The percentage error against the correct $2.96\ \text{cm}^3$ is $\dfrac{2.96 - 1.96}{2.96} \times 100 = \dfrac{1.00}{2.96} \times 100 = 33.8\ \%$ too small.
Check the correct answer for physical sense. The absolute pressure fell from $2.96$ to $1.00$ atmospheres, a factor of $2.96$ , so at fixed temperature the volume must grow by that same factor. It did: $1.00 \times 2.96 = 2.96\ \text{cm}^3$ . The mixed answer implies the pressure fell by a factor of only $1.96$ , which is the pressure ratio you would get if the atmosphere did not exist.

(a) $V_2 = 2.96\ \text{cm}^3$ . (b) Gauge pressures give a division by zero, because the gauge pressure at the surface is zero. (c) Mixing the scales gives $1.96\ \text{cm}^3$ , which is $33.8\ \%$ too small and carries no warning that anything went wrong.

Gas under a loaded piston, where the gas law meets the atmosphere

A vertical cylinder of cross-sectional area $0.010\ \text{m}^2$ is closed at the top by a freely sliding piston of mass $5.0\ \text{kg}$ , with the atmosphere above it. The gas below occupies $2.0 \times 10^{-3}\ \text{m}^3$ at $300\ \text{K}$ . Take $R = 8.31\ \text{J/(mol} \cdot \text{K)}$ , $g = 9.8\ \text{m/s}^2$ and $P_0 = 1.0 \times 10^5\ \text{Pa}$ . (a) Find the gauge pressure of the gas. (b) Find its absolute pressure. (c) Find the number of moles. (d) Find what a student gets by using the gauge pressure in the gas law instead.

(a) The piston is in equilibrium, so the gas below pushes up with exactly the force the piston's weight and the atmosphere above push down. The excess of the gas pressure over atmospheric is therefore whatever supports the piston's weight: $P_{\text{gauge}} = \dfrac{mg}{A}$ .
$mg = (5.0)(9.8) = 49\ \text{N}$ , and $P_{\text{gauge}} = 49 / 0.010 = 4900\ \text{Pa}$ .
(b) The absolute pressure adds the reference pressure sitting on the piston: $P = P_0 + \dfrac{mg}{A} = 1.0 \times 10^5 + 4900 = 1.049 \times 10^5\ \text{Pa}$ .
Note how small the piston's contribution is. It raises the pressure by under $5\ \%$ , which is precisely what makes the error in part (d) so easy to make and so large.
(c) The ideal gas law printed on the AP Physics 2 sheet is $PV = nRT$ , and its $P$ is the absolute pressure of the gas. So $n = \dfrac{PV}{RT} = \dfrac{(1.049 \times 10^5)(2.0 \times 10^{-3})}{(8.31)(300)}$ .
Numerator: $(1.049 \times 10^5)(2.0 \times 10^{-3}) = 209.8\ \text{J}$ . Denominator: $(8.31)(300) = 2493\ \text{J/mol}$ .
$n = 209.8 / 2493 = 0.08416\ \text{mol}$ , which is $0.084\ \text{mol}$ to two significant figures.
(d) Using the gauge pressure instead: numerator $(4900)(2.0 \times 10^{-3}) = 9.8\ \text{J}$ , so $n = 9.8 / 2493 = 0.003931\ \text{mol}$ , or $3.9 \times 10^{-3}\ \text{mol}$ .
That is wrong by a factor of $0.08416 / 0.003931 = 21.4$ . A pressure error of under $5\ \%$ became a mole-count error of more than twenty times, because the gauge value discarded almost the entire pressure rather than a small part of it.
The general shape of the trap: when the absolute pressure is dominated by the atmosphere, the gauge pressure is a small residue, and using it in place of the absolute value does not shift the answer slightly. It changes the order of magnitude.

(a) $P_{\text{gauge}} = 4.9 \times 10^3\ \text{Pa}$ . (b) $P = 1.049 \times 10^5\ \text{Pa}$ . (c) $n = 0.084\ \text{mol}$ . (d) The gauge pressure gives $3.9 \times 10^{-3}\ \text{mol}$ , low by a factor of $21.4$ , even though the two pressures differ by less than five percent.

Frequently asked questions

What is the difference between gauge pressure and absolute pressure?

They are the same physical pressure measured from two different zeros. Absolute pressure counts from a perfect vacuum and is the total push the fluid exerts at a point. Gauge pressure counts from the local atmosphere and is only the excess over it. AP Physics 1 essential knowledge 8.2.B.2 states that the absolute pressure of a fluid at a given point is equal to the sum of a reference pressure, such as the atmospheric pressure, and the gauge pressure, and the sheet prints that as P equals P nought plus rho g h. Essential knowledge 8.2.B.3 gives the gauge pressure of a vertical column of fluid on its own as rho g h. At sea level the two differ by one atmosphere, which the equation sheet gives as 1.0 times 10 to the fifth pascals.

Is rho g h the gauge pressure or the absolute pressure?

By itself, rho g h is the gauge pressure. AP Physics 1 essential knowledge 8.2.B.3 says exactly that: the gauge pressure of a vertical column of fluid is described by P gauge equals rho g h. To get the absolute pressure you add the reference pressure sitting on top of the column, which gives the sheet's other printed equation, P equals P nought plus rho g h. When the surface is open to the air that reference pressure is atmospheric, about 1.0 times 10 to the fifth pascals. When the space above the liquid has been evacuated, the reference pressure is zero and the two agree.

Do I use gauge or absolute pressure in the ideal gas law?

Absolute pressure, always. The P in PV equals nRT is the actual pressure of the gas, measured from a vacuum, and a scale whose zero sits at one atmosphere gives the wrong answer for the same reason Celsius gives the wrong answer in a formula that needs kelvin. The size of the error can be startling. A gas at a gauge pressure of 4900 pascals has an absolute pressure of 104,900 pascals, a difference of under five percent, but substituting the gauge value into the gas law understates the number of moles by a factor of more than twenty, because it throws away almost all of the real pressure rather than a small slice of it.

What value of atmospheric pressure does the AP equation sheet give?

The constants box at the top of the Table of Information prints 1 atm equals 1.0 times 10 to the fifth newtons per square meter, equals 1.0 times 10 to the fifth pascals. It appears on both the AP Physics 1 and the AP Physics 2 sheets. Two significant figures, so it is not the 101,325 pascals quoted in some chemistry courses, and an answer built on it cannot honestly carry more than two significant figures from that term. On the AP Physics 1 sheet the constants box holds only four lines in total: the universal gravitational constant, this atmosphere line, and the acceleration due to gravity written twice, once as 9.8 meters per second squared and once as 9.8 newtons per kilogram.

Can gauge pressure be negative?

Yes. Gauge pressure is the amount by which a pressure exceeds the surrounding atmosphere, so anything below atmospheric pressure has a negative gauge pressure. A partial vacuum in a sealed flask, or the inside of a drinking straw while you are sucking on it, both read negative on a gauge. Absolute pressure cannot go below zero, because zero absolute pressure is a perfect vacuum and there is nothing further to remove. This asymmetry is a quick way to tell which quantity a problem has given you: a negative pressure is a gauge pressure.

Does the buoyant force depend on gauge or absolute pressure?

Either one gives the same answer, because the buoyant force comes from a pressure difference between the bottom and the top of an object and the reference pressure cancels. Work a submerged cube of side 0.20 meters with its top 1.00 meter down in water. Using absolute pressures, the bottom face is pushed up with 4470.4 newtons and the top face pushed down with 4392 newtons, a net 78.4 newtons upward. Using gauge pressures, the two forces are 470.4 and 392 newtons, and the net is the same 78.4 newtons. The sheet's F sub b equals rho V g gives 78.4 newtons as well. This cancellation is why most of a fluids unit can be worked in gauge pressure without anyone noticing.

Why does a tire gauge read zero on a flat tire that still has air in it?

Because a tire gauge reports gauge pressure, and a flat tire is at atmospheric pressure, so it has zero excess over its surroundings. The air inside is still at an absolute pressure of about 1.0 times 10 to the fifth pascals. The same is true of every ordinary pressure instrument: a blood pressure cuff, a scuba tank dial, and a manometer all sit with the atmosphere pressing on one side of the sensing element, so what they measure is the difference. A tire inflated to a gauge reading of 2.2 times 10 to the fifth pascals holds air at an absolute 3.2 times 10 to the fifth pascals.