Pressure in Physics — Formulas, Examples & 12 Practice Problems
Pressure in Physics — Formulas, Examples & 12 Practice Problems
Pressure in Physics — Formulas, Examples and Practice Problems
12 fully worked problems covering the pressure formula P = F/A, hydrostatic pressure P = ρgh, absolute vs gauge pressure, Pascal’s principle and real-world applications
DEFINITION: Pressure = Force per unit area
P = F / AUNIT: Pascal (Pa) = 1 N/m²
KEY FORMULAS:
Basic: P = F / A [force ÷ area]
Hydrostatic: P = ρgh [fluid pressure at depth]
Absolute: P_abs = P_gauge + P_atm
Pressure head: h = P / (ρg)COMMON UNITS:
1 atm = 101,325 Pa = 14.696 psi = 1.01325 bar = 760 mmHg
RULES FOR P = F/A:
• F must be PERPENDICULAR to the surface
• A is the CONTACT area (not total surface area)
• Same force, smaller area → LARGER pressure
• Same force, larger area → SMALLER pressure
RULES FOR P = ρgh:
• ρ = fluid density (kg/m³)
• g = 9.807 m/s²
• h = depth below surface (m)
• Pressure increases linearly with depth
• Every 10 m of water ≈ +1 atm gauge pressure
What Is Pressure? — Definition and Formula
Pressure is the force exerted per unit area on a surface. The pressure formula is:
P = F / A
Where P is pressure in Pascals (Pa), F is force in Newtons (N), and A is the contact area in square metres (m²). One Pascal equals one Newton per square metre: 1 Pa = 1 N/m².
The key insight behind the pressure formula is that pressure depends on how the force is distributed. The same 100 N force creates very different pressures depending on the area it acts on:
Very HIGH pressure on a needle tip (area ≈ 10⁻⁶ m²) → P ≈ 10⁸ Pa = 100 MPa
Very LOW pressure on snowshoes (area ≈ 0.1 m²) → P ≈ 1,000 Pa = 1 kPa
This is why sharp knives cut easily, high heels damage floors, and snowshoes prevent sinking into snow — the same force, very different areas, vastly different pressures.
The pressure formula P = F/A can be rearranged to find any unknown: F = P × A (to find force) or A = F/P (to find area). This flexibility makes the pressure equation applicable to hydraulic systems, fluid mechanics, structural engineering and everyday physics problems.
Use the Pressure Calculator to solve for P, F, or A instantly with full working
These problems apply the pressure formula P = F/A directly. Focus on correct unit conversion — especially converting cm² to m² — before substituting into the pressure equation.
1
Book Resting on a Table — Basic Pressure Calculation
A book weighing 15 N rests on a table. The book’s cover has dimensions 20 cm × 25 cm. Calculate the pressure the book exerts on the table.
Step 1 — Identify known values
F = 15 N (weight of book = gravitational force downward)
A = 20 cm × 25 cm
= 0.20 m × 0.25 m ← convert cm to m
= 0.05 m²
Step 2 — Apply the pressure formula
P = F / A
Step 3 — Substitute values
P = 15 N / 0.05 m²
Step 4 — Calculate
P = 300 Pa = 300 N/m²
Step 5 — Convert to common units
P = 300 Pa = 0.300 kPa = 0.0435 psi
✓ Result: P = 300 Pa
Context: 300 Pa is much less than atmospheric pressure (101,325 Pa). Atmospheric pressure acts equally on all surfaces, so only the additional pressure from the book matters structurally. This is why books don’t crush tables — the pressure is negligible compared to what the table is already experiencing from the atmosphere.
2
High Heels vs Flat Shoes — The Pressure Difference That Damages Floors
A woman wearing high heels (total mass 60 kg) balances on one heel. The heel has a contact area of 1 cm². Calculate the pressure under the heel. Compare this to a flat shoe with contact area 150 cm².
Step 1 — Calculate force (weight)
F = mg = 60 kg × 9.807 m/s² = 588.4 N
Step 2 — High heel calculation
A_heel = 1 cm² = 1 × 10⁻⁴ m² ← 1 cm² = 10⁻⁴ m² (NOT 10⁻² m²)
P_heel = F / A = 588.4 / (1 × 10⁻⁴)
= 5,884,000 Pa = 5.88 MPa = 853 psi
Step 3 — Flat shoe calculation
A_flat = 150 cm² = 150 × 10⁻⁴ m² = 0.015 m²
P_flat = 588.4 / 0.015
= 39,227 Pa = 39.2 kPa = 5.69 psi
Step 4 — Compare pressures
Ratio = P_heel / P_flat = 5,884,000 / 39,227 = 150
The high heel exerts 150× MORE pressure than the flat shoe.
⚠ Common Mistake Avoided — cm² to m² conversion
1 cm² = 10⁻⁴ m² (NOT 10⁻² m²). If you divide by 0.01 instead of 0.0001, you get 58,840 Pa — 100× too low. Always multiply cm² by 10⁻⁴ to get m².
Why this matters: This is why high heels damage hardwood floors and sink into soft ground — the pressure formula P = F/A shows that reducing area by 150× multiplies pressure by 150×, even though the total force (body weight) is identical.
Rearranging the Pressure Formula — Finding F and A
The pressure formula P = F/A can be rearranged to find force (F = P × A) or area (A = F/P). These rearrangements are essential for engineering design problems where you know the pressure constraint and need to determine the required geometry or the resulting force.
3
Hydraulic Press — Finding Minimum Piston Area (A = F/P)
A hydraulic press must exert a force of 50,000 N on a component. The maximum pressure the system can generate is 20 MPa. What minimum area must the piston have?
Step 1 — Rearrange the pressure formula to find area
P = F / A → A = F / P
Step 2 — Substitute known values
A = 50,000 N / 20,000,000 Pa
A = 0.0025 m² = 25 cm²
A = πr² → r = √(A/π)
r = √(0.0025 / π)
r = √(7.958 × 10⁻⁴)
r = 0.0282 m = 2.82 cm
Minimum diameter = 2r = 5.64 cm
✓ Minimum piston area = 25 cm² | Circular piston: diameter ≥ 5.64 cm
4
Force on a Pressure Vessel Wall (F = P × A)
A pressure of 150 kPa acts on a rectangular surface of dimensions 40 cm × 60 cm. What is the total force on the surface?
Step 1 — Rearrange the pressure formula to find force
P = F / A → F = P × A
Step 2 — Calculate area
A = 0.40 m × 0.60 m = 0.24 m²
Step 3 — Substitute and calculate
F = 150,000 Pa × 0.24 m²
F = 36,000 N = 36 kN
In other units:
36,000 / 9.807 = 3,670 kgf
36,000 / 4.448 = 8,093 lbf ≈ 3.67 tonnes of force
✓ Total force = 36,000 N = 36 kN ≈ 3.67 tonnes of force
Design implication: Even moderate pressures create enormous forces over large areas. A tank wall measuring just 40 cm × 60 cm at 150 kPa (roughly 1.5 atmospheres) experiences nearly 4 tonnes of force. This is why pressure vessels require thick walls, high-strength materials, and careful engineering — and why understanding the pressure formula P = F/A is fundamental to mechanical design.
🔄 Need to convert pressure units? Pa → psi → bar → atm instantly
In a fluid at rest, the hydrostatic pressure at depth h below the surface is given by:
P = ρgh
where ρ is fluid density (kg/m³), g is gravitational acceleration (9.807 m/s²), and h is depth (m). This is gauge pressure — the pressure above atmospheric. For total absolute pressure, add P_atm = 101,325 Pa. Hydrostatic pressure increases linearly with depth; every additional metre of water adds ρg Pa of pressure.
5
Swimming Pool — Gauge and Absolute Hydrostatic Pressure
Calculate the water pressure at the bottom of a swimming pool 3 metres deep. Give both gauge pressure and total absolute pressure.
Step 1 — Identify values
ρ = 1000 kg/m³ (fresh water)
g = 9.807 m/s²
h = 3 m
Step 2 — Calculate gauge pressure using hydrostatic pressure formula P = ρgh
Submarine at 300 m Depth — Hull Pressure in Seawater
A submarine operates at a depth of 300 m in seawater (ρ = 1,025 kg/m³). Calculate the total pressure the hull must withstand. Express in Pa, bar, and psi.
⚠ Common Mistake Avoided — seawater vs fresh water density
Using fresh water density (1000 kg/m³) instead of seawater (1025 kg/m³) gives 3,015,000 − 94,500 = a 2.5% lower answer. For accurate hydrostatic pressure calculations, always use the correct fluid density: fresh water = 1000, seawater = 1025, mercury = 13,534 kg/m³.
7
Depth for 1 atm Gauge Pressure — Deriving the “10 m Rule”
At what depth in fresh water does the gauge pressure equal exactly 1 atm (101,325 Pa)?
Step 1 — Rearrange the hydrostatic pressure formula for depth
P = ρgh → h = P / (ρg)
Step 2 — Substitute values
h = 101,325 / (1000 × 9.807)
h = 101,325 / 9,807
h = 10.332 m
Step 3 — Seawater comparison
For seawater (ρ = 1025 kg/m³):
h = 101,325 / (1025 × 9.807)
h = 101,325 / 10,052
h = 10.08 m
✓ h = 10.33 m in fresh water | h = 10.08 m in seawater
Origin of the “10 m rule”: The hydrostatic pressure formula P = ρgh shows that the exact depth for 1 atm gauge pressure is 10.33 m in fresh water (not exactly 10 m). The “every 10 m of water ≈ 1 atm” rule is a rounded approximation. For seawater, the exact depth is actually slightly less (10.08 m) because seawater is denser — a critical distinction for scuba diving depth calculations.
Gauge pressure measures pressure relative to atmospheric pressure. Absolute pressure measures pressure relative to a perfect vacuum. The relationship is: P_absolute = P_gauge + P_atmospheric. At sea level, P_atm = 101,325 Pa = 14.696 psi = 1.01325 bar. Getting this distinction wrong is one of the most common errors in pressure problems.
8
Car Tyre — Converting Gauge Pressure to Absolute Pressure
A car tyre has a gauge pressure of 32 psi. What is the absolute pressure in the tyre in Pa, psi, and bar?
Step 1 — Convert gauge pressure to Pa
P_gauge = 32 psi × 6,894.757 Pa/psi = 220,632 Pa
Step 2 — Add atmospheric pressure to get absolute pressure
Practical note: “32 psi” on a tyre gauge always means 32 psig (gauge). The actual air inside the tyre is at 46.7 psia. A completely flat tyre still has 14.7 psia — it’s at atmospheric pressure, not zero absolute pressure. This distinction between gauge and absolute pressure matters everywhere from tyre maintenance to industrial process control.
9
Flat Tyre at Altitude — Why Gauge Readings Change
A pressure gauge reads 0 psi for a fully deflated tyre at sea level. What is the absolute pressure? If you drive to an altitude where atmospheric pressure is 85 kPa, what does the gauge read now for the same tyre (assume rigid tyre, same air inside)?
✓ The “flat” tyre shows 2.37 psi of gauge pressure at altitude!
Insight: The absolute pressure inside the tyre doesn’t change (same air, rigid container), but the atmospheric reference against which gauge pressure is measured decreases. This is why tyres appear to gain pressure when you drive up a mountain, and why tyre pressure should always be checked at the same altitude. It is also why the gauge pressure / absolute pressure distinction is not merely academic — it has direct practical consequences for measurements.
🔁 Convert between absolute and gauge pressure instantly
Pressure and Force Problems — Real-World Applications
The pressure formula P = F/A appears across engineering, biology and everyday life. These problems demonstrate why understanding the relationship between pressure, force, and area matters far beyond the classroom.
10
⭐ Elephant vs High Heel — The Most Counterintuitive Pressure Problem in Physics
An elephant has a mass of 5,000 kg. Its four feet each have an approximate contact area of 0.05 m². Compare the pressure an elephant exerts on the ground to:
(a) A 60 kg woman standing on both feet (each foot: 150 cm²)
(b) A 60 kg woman balancing on one high heel (contact area: 1 cm²)
Elephant — applying the pressure formula P = F/A
F_elephant = 5,000 × 9.807 = 49,035 N
A_total = 4 × 0.05 = 0.20 m²
P_elephant = 49,035 / 0.20
= 245,175 Pa = 245.2 kPa = 35.6 psi
(a) Woman on both flat feet — pressure formula P = F/A
F_woman = 60 × 9.807 = 588.4 N
A = 2 × 150 cm² = 300 cm² = 300 × 10⁻⁴ = 0.030 m²
P_flat = 588.4 / 0.030
= 19,613 Pa = 19.6 kPa = 2.85 psi
(b) Woman on one high heel — pressure formula P = F/A
F_woman = 588.4 N (same body weight)
A_heel = 1 cm² = 1 × 10⁻⁴ m²
P_heel = 588.4 / (1 × 10⁻⁴)
= 5,884,000 Pa = 5.88 MPa = 853 psi
Comparison — the counterintuitive result
Elephant: 35.6 psi (large feet distribute enormous weight)
Woman flat: 2.85 psi (very low — wide contact area)
Woman heel: 853 psi (24× HIGHER than the elephant!)
Ratio (heel vs elephant) = 853 / 35.6 = 24×
✓ Woman on high heel: 853 psi | Elephant: 35.6 psi
✓ The 60 kg woman on a single stiletto exerts 24 TIMES more pressure than a 5,000 kg elephant.
Why this result seems impossible — and why it’s correct:
The pressure formula P = F/A makes it mathematically inevitable. The elephant is heavier (49,035 N vs 588 N — 83× more force), but its feet are enormously wider (0.20 m² vs 10⁻⁴ m² — 2000× more area). The area advantage outweighs the force difference by a factor of 24. This is the single most powerful demonstration that pressure depends on BOTH force AND area, not force alone. Students who encounter this comparison remember the pressure formula permanently — because it resolves what initially seems physically impossible. This is why high heels destroy hardwood floors, sink into lawns, and are banned from many historic buildings, while elephants roam savannah and forest floors without leaving deep imprints.
11
Hydraulic Jack — Pascal’s Principle and Mechanical Advantage
A hydraulic jack has a small piston of area 5 cm² and a large piston of area 500 cm². A force of 200 N is applied to the small piston. What force is exerted by the large piston? (Pascal’s Principle)
Pascal’s Principle — pressure is transmitted equally throughout the fluid
P_small = P_large
(Pressure applied at one point in an enclosed fluid
is transmitted undiminished in all directions)
Step 1 — Pressure on small piston using P = F/A
A₁ = 5 cm² = 5 × 10⁻⁴ m² ← 1 cm² = 10⁻⁴ m²
P = F₁ / A₁ = 200 / (5 × 10⁻⁴)
P = 400,000 Pa
Step 2 — Force on large piston (same pressure, larger area)
A₂ = 500 cm² = 500 × 10⁻⁴ m² = 0.05 m² ← 500 × 10⁻⁴ (NOT 500 × 10⁻²)
F₂ = P × A₂ = 400,000 × 0.05
F₂ = 20,000 N = 20 kN = 2,039 kgf
✓ Output force = 20,000 N = 20 kN | Mechanical advantage = 100×
Application: A 200 N hand force (≈ 20 kg) lifts 20,000 N (≈ 2 tonnes). This is Pascal’s Principle — the foundation of all hydraulic machinery: car lifts, excavators, hydraulic brakes, aircraft control surfaces, and injection moulding machines. The pressure formula P = F/A is the engine driving all of these systems.
⚠ Common Mistake Avoided — 500 cm² to m² conversion
500 cm² = 500 × 10⁻⁴ m² = 0.05 m² (NOT 500 × 10⁻² = 5 m²). Using 5 m² instead of 0.05 m² gives F₂ = 2,000,000 N — 100× too high. Verify: 1 cm² = 10⁻⁴ m² throughout all pressure calculations.
Challenge Problem — Multi-Step Pressure
This problem combines the pressure formula P = F/A, the hydrostatic pressure formula P = ρgh, gauge vs absolute pressure, and integral calculus to verify the centre-pressure approximation. It represents the level of multi-concept pressure problems found in university-level fluid mechanics courses.
12
Oil Tank with Submerged Plate — Full Multi-Step Analysis
A rectangular tank (2 m × 3 m base, 4 m tall) is filled with oil (ρ = 850 kg/m³) to a depth of 3.5 m. A flat plate (0.5 m × 0.5 m) is mounted on the side wall with its centre at depth 2 m.
(a) Gauge pressure at the centre of the plate
(b) Total force on the plate due to oil pressure
(c) Total absolute pressure at the bottom of the tank
Given values
Oil density: ρ = 850 kg/m³
Fill depth: 3.5 m
Plate centre: h = 2 m below oil surface
Plate: 0.5 m × 0.5 m (plate edges: depth 1.75 m to 2.25 m)
g: 9.807 m/s²
(a) Gauge pressure at plate centre — using hydrostatic formula P = ρgh
✓ (a) Gauge pressure at plate centre: 16.67 kPa = 2.42 psi
✓ (b) Force on plate: 4,168 N ≈ 4.17 kN = 937 lbf
✓ (c) Absolute pressure at bottom: 130.5 kPa = 1.29 atm
Pressure in Everyday Life — Why It Matters
The pressure formula P = F/A and the hydrostatic pressure formula P = ρgh govern phenomena across an enormous range of scales — from blood pressure in the circulatory system to the crushing pressures of the deep ocean. The table below shows real-world gauge and absolute pressures alongside atmospheric pressure as a reference baseline.
Situation
Pressure
Notes
Atmospheric pressure (sea level)
101,325 Pa = 14.7 psi absolute
Acts on all surfaces equally; reference for gauge pressure
Human blood pressure (systolic)
~16 kPa gauge = 120 mmHg
Gauge pressure above atmospheric; measured in mmHg by convention
Car tyre (inflated)
~230 kPa gauge = 33 psi gauge
Tyre gauge reads gauge pressure; absolute ≈ 330 kPa
Scuba depth limit — recreational (30 m)
~400 kPa absolute = 4 atm
Hydrostatic pressure P = ρgh at 30 m seawater
Bicycle tyre (road racing)
~700 kPa gauge = 100 psi gauge
Far higher than car tyres — thin, hard tyres minimise rolling resistance
Jet aircraft cabin
~75 kPa absolute
Pressurised to equivalent of ~2,500 m altitude; lower than sea level
The pressure formula P = F/A can be visualised as a triangle. Cover the variable you want to find and the remaining two show you how to calculate it. All three rearrangements follow directly from the single pressure equation.
Cover P → find Pressure
P = F / A
Example: F = 500 N, A = 0.02 m² P = 500 / 0.02 = 25,000 Pa
Cover F → find Force
F = P × A
Example: P = 300 kPa, A = 0.5 m² F = 300,000 × 0.5 = 150,000 N
Cover A → find Area
A = F / P
Example: F = 800 N, P = 40,000 Pa A = 800 / 40,000 = 0.02 m²
Key rule: In all three rearrangements of the pressure formula, F must be the component of force perpendicular to the surface A. If a force acts at an angle θ to the surface normal, use F_⊥ = F × cos(θ) before applying P = F/A.
Common Mistakes in Pressure Problems
The five errors below account for the majority of wrong answers in pressure calculations. Recognising these mistakes before an exam is the fastest way to improve your score on pressure formula problems.
Mistake 1 — Using total surface area instead of contact area
A brick lying flat on a table has a different contact area than the same brick standing on its end. The weight (force) is identical, but the pressure formula P = F/A requires only the contact area between the brick and the table — not the total surface area of the brick. Using the wrong area changes the calculated pressure by a large factor.
Mistake 2 — Wrong cm² to m² conversion (most common unit error)
1 cm² = 10⁻⁴ m² (NOT 10⁻² m²). This is because 1 cm = 10⁻² m, so 1 cm² = (10⁻² m)² = 10⁻⁴ m². Dividing by 0.01 instead of 0.0001 gives a pressure answer 100× too small. Always multiply cm² by 10⁻⁴ to convert to m² before applying the pressure formula.
Mistake 3 — Mixing gauge pressure and absolute pressure
A tyre gauge reading of “32 psi” is gauge pressure (32 psig). Absolute pressure is 32 + 14.7 = 46.7 psia. The hydrostatic pressure formula P = ρgh gives gauge pressure only — add atmospheric pressure (101,325 Pa) to get absolute pressure. Using gauge pressure where absolute is needed (e.g., in the ideal gas law) produces significant errors.
Mistake 4 — Wrong fluid density in hydrostatic pressure P = ρgh
Common densities: fresh water = 1,000 kg/m³; seawater = 1,025 kg/m³; mercury = 13,534 kg/m³; oil ≈ 800–900 kg/m³. Using fresh water density for seawater in the hydrostatic pressure formula gives a result 2.5% too low. Using water density for mercury gives a result 13.5× too low.
Mistake 5 — Applying P = F/A to non-perpendicular forces
The pressure formula P = F/A requires F to be perpendicular to the surface A. If a force F acts at angle θ to the surface normal, only the perpendicular component contributes to pressure: F_⊥ = F cos(θ). Forgetting to resolve the force overestimates the pressure when the force is not perfectly perpendicular to the surface.
Pressure and Volume Relationship — Boyle’s Law
For a fixed amount of gas at constant temperature, pressure and volume are inversely proportional. This is Boyle’s Law — one of the fundamental gas laws and the reason the pressure and volume relationship appears in scuba diving physics, syringe mechanics, and pneumatic engineering. The pressure and volume relationship is expressed as:
P₁V₁ = P₂V₂
When pressure doubles, volume halves. When pressure halves, volume doubles. The graph of pressure versus volume is a hyperbola — pressure and volume of a gas vary inversely along a constant-temperature curve called an isotherm. How are the pressure and volume of a gas related? They are inversely proportional at constant temperature, which is the defining statement of Boyle’s Law.
Example 1 — Scuba tank: A diver breathes air at 4 atm absolute (30 m depth). At the surface (1 atm), the same mass of air occupies 4× the volume. This is why ascent must be slow — air in the lungs expands and must be exhaled continuously to avoid barotrauma.
Example 2 — Syringe: A syringe contains 20 mL of air at 101 kPa. The plunger is pushed until the volume is 5 mL. What is the new pressure (temperature constant)?
The pressure formula is P = F / A, where P is pressure in Pascals (Pa), F is the perpendicular force in Newtons (N), and A is the contact area in square metres (m²). For fluid pressure at depth, the hydrostatic pressure formula is P = ρgh, where ρ is fluid density (kg/m³), g = 9.807 m/s², and h is depth below the surface (m). The pressure equation can be rearranged to find force (F = P × A) or area (A = F/P).
What is hydrostatic pressure and when does P = ρgh apply?
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid column above a given point. The hydrostatic pressure formula P = ρgh applies whenever a fluid is static (not flowing) and you need the pressure at a specific depth. It gives gauge pressure — the pressure above atmospheric. For absolute total pressure, add atmospheric pressure: P_abs = ρgh + P_atm. Hydrostatic pressure increases linearly with depth; every 10.33 m of fresh water (or 10.08 m of seawater) adds one atmosphere of gauge pressure.
What is the difference between gauge pressure and absolute pressure?
Gauge pressure is measured relative to atmospheric pressure (what a tyre gauge or pressure gauge reads). Absolute pressure is measured relative to a perfect vacuum and includes atmospheric pressure. The conversion formula is: P_absolute = P_gauge + P_atmospheric. At sea level, P_atm = 101,325 Pa = 14.696 psi = 1 atm. A tyre reading 32 psi gauge (32 psig) has an absolute pressure of 32 + 14.7 = 46.7 psia. A completely flat tyre still has 14.7 psia absolute — it is at atmospheric pressure, not zero.
What is Pascal’s Principle?
Pascal’s Principle states that pressure applied to an enclosed, incompressible fluid is transmitted equally and undiminished in all directions throughout the fluid. The practical consequence: in a hydraulic system with a small piston (area A₁) and large piston (area A₂), if force F₁ is applied to the small piston, the pressure P = F₁/A₁ is transmitted to the large piston, producing output force F₂ = P × A₂ = F₁ × (A₂/A₁). The mechanical advantage equals the ratio of piston areas. This is the operating principle behind car lifts, hydraulic jacks, excavators, and hydraulic brakes.
Why does pressure increase with depth in a fluid?
Pressure increases with depth because each additional layer of fluid must support the weight of all the fluid above it. The hydrostatic pressure formula P = ρgh shows that pressure at depth h equals the weight per unit area of the fluid column above: ρ (mass per volume) × g (gravitational acceleration) × h (column height). The deeper you go, the taller the fluid column above, the greater its weight, and the greater the pressure it exerts. This is why submarines require enormously strong pressure hulls, and why the pressure at the Mariana Trench (11 km depth) reaches about 1,070 atmospheres.
How do you convert gauge pressure to absolute pressure?
Add atmospheric pressure to gauge pressure: P_absolute = P_gauge + P_atmospheric. At sea level: P_atm = 101,325 Pa = 14.696 psi = 1.01325 bar = 1 atm. To convert the other direction (absolute to gauge): P_gauge = P_absolute − P_atmospheric. Important: the ideal gas law and Boyle’s Law (P₁V₁ = P₂V₂) require absolute pressure. Using gauge pressure in these equations produces incorrect results.
