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With gauge pressure $p_g$ referenced to atmospheric pressure $p_{\mathrm{atm}}$, absolute pressure is $p=p_g+p_{\mathrm{atm}}$. Thus Boyle's law becomesEven a constant atmospheric pressure cannot be dropped from these sums. For the calculated $100\mathrm{kPa}$ to $200\mathrm{kPa}$ example, take $p_{\mathrm{atm}}=100\mathrm{kPa}$. The initial and final gauge pressures are $0$ and $100\mathrm{kPa}$; their products with the respective volume are $0$ and $5000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3$, which are unequal. Zero gauge pressure means balance with the surroundings, not absence of gas or absence of wall momentum transfer.

For a frictionless piston of surface area $A$ with its outer face exposed to atmospheric pressure, the additional inward force required for mechanical equilibrium is $F=(p-p_{\mathrm{atm}})A=p_{g}A$. This is why a mechanical load can reveal a pressure difference even though the equation of state requires absolute pressure. With piston friction, the load alone no longer determines $p_{g}A$; a direct pressure sensor avoids that particular inference.

Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

Vernier's Boyle's Law: Pressure-Volume Relationship in Gases, Experiment 6 of Chemistry with Vernier, specifies a graduated $20\mathrm{mL}$ gas syringe, a pressure sensor and LabQuest recording. Preparation sets the disconnected syringe to $10.0\mathrm{mL}$, then attaches it to the sensor valve, trapping air. At several plunger positions, the displayed pressure is allowed to stabilize before the paired volume is entered. The analysis tests whether $pV$ remains approximately constant. It adds $0.8\mathrm{mL}$ to each syringe reading for the sensor's internal volume; this correction belongs to that apparatus. The worksheet supplies a blank results table, not measured values.^[ref]

The manufacturer's Gas Pressure Sensor User Manual identifies a membrane sensor referenced to a vacuum, hence an absolute pressure readout. Its calibration section uses the actual local atmospheric pressure for an ambient reference check, rather than a weather report corrected to sea level. Checking this baseline before sealing separates an offset error from compression physics.^[ref]

Additional controls follow from the model. Keep the same sealed amount of substance, connections and ambient temperature; persistent pressure drift at held volume after thermal equilibrium suggests a leak or another uncontrolled change. Small steps and waiting allow compression heat to leave the gas; avoiding hand warming reduces another heat input. Compare settled readings when approaching a position from either direction. Plunger friction affects the applied force and position reproducibility, so record actual pressure and actual plunger position rather than predicting pressure from the load. These are proposed controls, not reported experimental outcomes.

Nearly constant $pV$ across equilibrated states supports the fixed-temperature ideal gas approximation over the tested range. Stabilized pressure alone does not demonstrate uniform or constant temperature; the isothermal gas syringe experiment adds a temperature readout to distinguish a transient from its cooled endpoint. Finite sensor accuracy and uncertain total volume limit the test, and agreement does not establish ideal behavior at every pressure. Repeated return readings help distinguish reproducibility from a one-direction trend.

A gas syringe makes the distinction between a constant-temperature path and equal-temperature endpoints observable. The Ideal Gas Law Apparatus (TD-8596A), Instruction Manual 012-08420B by PASCO scientific specifies a TD-8596A gas syringe with a mechanical stop, an internal thermistor, an attached pressure sensor and data acquisition. Record the stop volume with the coupling disconnected, then set $40\mathrm{mL}$ and seal the coupling. Record pressure and temperature, compress quickly to the stop near $20\mathrm{mL}$, and hold until both readouts settle. The manual gives roughly $30\mathrm{s}$ for settling and about $0.5\mathrm{s}$ for the thermistor response. Releasing the plunger gives a second transient and a final volume to record.^[ref]

The manual describes a pressure overshoot and temperature rise during rapid compression, followed by cooling at the held volume; the settled pressure remains higher than initially. This illustrates approximately adiabatic compression followed by heat rejection, rather than an isothermal compression throughout. Use the settled initial and compressed readings to test Boyle's law, correcting both syringe readings by the unmarked tubing volume $V_d$:Use absolute pressure, not gauge pressure. A transient comparison of $pV/T$ must pair simultaneous pressure and temperature readings: the sensor lag can separate their peaks.^[ref]

For an additional test of the path, repeat with small compression steps, wait for stable temperature and pressure at every step, and repeat in reverse. Keep the sealed amount of substance, connecting tubing and ambient temperature fixed; prevent hand warming of the barrel and check for leaks by watching for continued pressure drift at held volume after thermal equilibrium. These are proposed controls, rather than reported measurements. Agreement of equilibrated points with Boyle's law supports the fixed-temperature ideal gas law approximation, but does not prove that the rapid segments between them were isothermal processes. A single local temperature sensor can miss temperature gradients; plunger friction can produce direction-dependent readings. The pressure and volume data estimate reversible work only insofar as the path approaches a reversible thermodynamic process; they do not directly measure heat. After motion stops, cooling at fixed volume has zero boundary work and $Q=\Delta U<0$, even though the complete equal-endpoint-temperature ideal gas run has $\Delta U=0$.

A linearized plot of Boyle's law uses $x=1/V$ as its horizontal coordinate. Then absolute pressure satisfies $p=Cx$: the slope is $C=nRT$, and the ideal intercept is zero. In the figure, $(V,p)=(100,100)$ and $(50,200)$ in $\mathrm{c}{\mathrm{m}}^3$ and $\mathrm{kPa}$ become $(x,p)=(0.010,100)$ and $(0.020,200)$ in $\mathrm{c}{\mathrm{m}}^{-3}$ and $\mathrm{kPa}$. Both give slope $10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3$. The Vernier experiment includes this reciprocal-volume graph as its extension.^[ref]

At constant atmospheric pressure, gauge pressure instead obeys $p_g=Cx-p_{\mathrm{atm}}$, with the same slope and a negative intercept. A nonzero fitted intercept in an intended absolute pressure plot can indicate sensor bias or an incorrect volume correction, but is not by itself a unique diagnosis. The origin is a mathematical extrapolation to infinite volume and zero absolute pressure, rather than an additional measured state.

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

A pressure measurement determines pressure relative to a specified reference using an instrument such as a manometer or pressure transducer. Gauge pressure, absolute pressure and differential pressure require different references; measurement calibration and measurement uncertainty determine how the reading can be interpreted.

A pressure sensor converts a pressure into a recorded signal. An absolute pressure sensor uses a vacuum reference; a gauge pressure sensor uses a specified reference pressure. A closed gas syringe and an absolute pressure sensor can test Boyle's law after allowing the temperature to stabilize.

A U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]