Standard Atmospheric Pressure, Applications, and Hydrostatic Pressure

Standard Atmospheric Pressure and Measurement

  • Definition of Standard Atmospheric Pressure

    • Standard atmospheric pressure (also known as normal atmospheric pressure) is defined as the atmospheric pressure measured at sea level.

    • It corresponds to a mercury column height of 760 mm Hg760\,\text{mm Hg} or 0.76 m0.76\,\text{m} (0.760 m0.760\,\text{m}).

    • This baseline standard pressure is designated as 1 atmosphere1\,\text{atmosphere} (1 atm1\,\text{atm}).

  • Mathematical Calculation of Atmospheric Pressure in Pascals

    • The atmospheric pressure is computed using the hydrostatic pressure formula p=ρghp = \rho g h.

    • Density of mercury (ρ\rho): 13590 kg m−313590\,\text{kg\,m}^{-3}.

    • Acceleration due to gravity (gg): 9.8 m s−29.8\,\text{m\,s}^{-2}.

    • Height of the mercury column (hh) at standard atmospheric pressure: 760 mm=0.760 m760\,\text{mm} = 0.760\,\text{m}.

    • Substitution and calculation:     Atmospheric pressure=ρgh=13590×9.8×0.760=1.013×105 Pa\text{Atmospheric pressure} = \rho g h = 13590 \times 9.8 \times 0.760 = 1.013 \times 10^5\,\text{Pa}

  • Standard Atmospheric Pressure Unit Conversions

    • 1 atm=1.013×105 Pa1\,\text{atm} = 1.013 \times 10^5\,\text{Pa}

    • 1 atm=1013 hPa1\,\text{atm} = 1013\,\text{hPa}

    • 1 atm=1.01 bar1\,\text{atm} = 1.01\,\text{bar}

    • 1 atm=14.7 lb in−2 (psi)1\,\text{atm} = 14.7\,\text{lb\,in}^{-2}\,(\text{psi})

    • 1 atm=760 mm Hg1\,\text{atm} = 760\,\text{mm Hg}

    • 1 atm=760 torr1\,\text{atm} = 760\,\text{torr}

  • Mechanics and Properties of the Mercury Barometer

    • The vertical height hh of the mercury column depends exclusively on the atmospheric pressure outside the tube.

    • Effect of Tilting: Tilting the barometer tube does not change the vertical height hh of the mercury column; it remains completely unaffected.

    • Effect of Tube Diameter: The vertical height hh is independent of the diameter or width of the tube.

    • Pressure Invariance: Pressures are equal at all points located along the same horizontal liquid level because pressure in a liquid does not depend on container angle or container width.

    • Lowered Tube Behavior: If the tube is lowered below a vertical height of 760 mm760\,\text{mm}, mercury completely fills the entire tube.

    • Torricellian Vacuum: The evacuated space created above the mercury column at the closed top end of the barometer tube is termed the Torricellian vacuum.

Practical Applications of Atmospheric Pressure

  • Sucking Liquid Through a Straw

    • Sucking action expands the volume of the lungs.

    • Expanding lung volume reduces air pressure within both the lungs and the mouth cavity.

    • External atmospheric pressure acting on the exposed surface of the liquid becomes greater than the reduced air pressure in the mouth.

    • This pressure differential forces the liquid to rise through the straw and into the mouth.

  • Drawing Liquid into a Syringe

    • Pulling the piston upward increases internal cylinder volume, causing pressure inside the cylinder to decrease.

    • External atmospheric pressure acting on the liquid surface drives the liquid through the nozzle and into the cylinder.

  • Pressing a Rubber Sucker on a Smooth Surface

    • Pressing the rubber sucker flat against a smooth surface squeezes out most of the air trapped inside.

    • Squeezing out air reduces the internal pressure beneath the sucker.

    • High external atmospheric pressure pressing against the outside surface holds the rubber sucker firmly in place.

Quantitative Examples and Reviewed Exercises

  • Example 3.2: Pressure Unit Conversion

    • Express 2 atm2\,\text{atm} pressure in mm Hg\text{mm Hg} and bars:

    • Conversion to mm Hg\text{mm Hg}:       2 atm=2×760 mm Hg=1520 mm Hg2\,\text{atm} = 2 \times 760\,\text{mm Hg} = 1520\,\text{mm Hg}

    • Conversion to bars:       2 atm=2×1.01 bar=2.02 bar2\,\text{atm} = 2 \times 1.01\,\text{bar} = 2.02\,\text{bar}

  • Example 3.3: Force Comparison Between Man and Child

    • Comparison of atmospheric pressures and total atmospheric forces acting on a man and a child standing side by side at the same location:

    • Let F1F_1 be the force acting on the man, F2F_2 be the force acting on the child, A1A_1 be the body surface area of the man, and A2A_2 be the body surface area of the child.

    • Surface area relationship: A_1 > A_2.

    • Because both individuals stand side by side at the exact same location, the atmospheric pressure pp acting on both is identical.

    • Utilizing force relation F=pAF = p A:       p A_1 > p A_2       F_1 > F_2

    • Conclusion: The total atmospheric force acting on the man is greater than the total atmospheric force acting on the child due to his larger body surface area.

  • Reviewed Exercise 1: Barometer Fluid Selection

    • Mercury is chosen over water in barometers despite mercury being a hazardous substance because the density of mercury is vastly greater than the density of water (\rho_{\text{Hg}} > \rho_{\text{H}_2\text{O}}).

    • The high density of mercury results in a manageable column height (760 mm760\,\text{mm} at standard pressure), whereas a water barometer would require a column height exceeding 10 m10\,\text{m}.

  • Reviewed Exercise 2: Barometric Altitude Calculation

    • Atmospheric pressure at sea level: 76 cm Hg=760 mm Hg76\,\text{cm Hg} = 760\,\text{mm Hg}.

    • Pressure drop rate: 10 mm Hg10\,\text{mm Hg} per 120 m120\,\text{m} of vertical ascent.

    • Mountain peak barometer reading: 70.5 cm Hg=705 mm Hg70.5\,\text{cm Hg} = 705\,\text{mm Hg}.

    • Pressure difference calculation:     Δp=760 mm Hg−705 mm Hg=55 mm Hg\Delta p = 760\,\text{mm Hg} - 705\,\text{mm Hg} = 55\,\text{mm Hg}

    • Mountain height calculation:     Height=55 mm Hg10 mm Hg×120 m=5.5×120 m=660 m\text{Height} = \frac{55\,\text{mm Hg}}{10\,\text{mm Hg}} \times 120\,\text{m} = 5.5 \times 120\,\text{m} = 660\,\text{m}

Pressure in Liquids and Hydrostatic Principles

  • Origin and Factors Influencing Liquid Pressure

    • A liquid exerts pressure on its container walls and bottom entirely because of its weight.

    • Liquid pressure depends directly on the depth beneath the liquid surface and the density of the liquid.

  • Derivation of Hydrostatic Pressure Equation

    • Consider a cylindrical container with a horizontal bottom surface area AA, filled with liquid of density ρ\rho up to depth/height hh:

    • Volume of liquid (VV):       V=AhV = A h

    • Mass of liquid (mm):       m=ρV=ρAhm = \rho V = \rho A h

    • Weight of liquid (ww):       w=mg=ρgAhw = m g = \rho g A h

    • Force exerted on the bottom of container (FF) equals the weight of liquid (ww):       F=w=ρgAhF = w = \rho g A h

    • Pressure exerted at the bottom surface (pp):       p=FA=wA=ρgAhAp = \frac{F}{A} = \frac{w}{A} = \frac{\rho g A h}{A}       p=ρghp = \rho g h

  • Characteristics of Liquid Pressure

    • Proportionality: Liquid pressure p=ρghp = \rho g h is directly proportional to depth hh and liquid density ρ\rho.

    • Independence of Area: Although total liquid weight depends on base area, liquid pressure pp is completely independent of the base area or cross-sectional area AA.

    • General Validity: The equation p=ρghp = \rho g h applies not only to the bottom surface of a container, but to any specific point or depth inside the liquid.

  • Total (True) Pressure at Depth

    • The term p=ρghp = \rho g h accounts solely for the liquid pressure itself.

    • When open to the environment, atmospheric pressure patmp_{\text{atm}} acts on the top surface of the liquid.

    • The true (total) pressure PP at depth hh inside liquid is:     P=patm+ρghP = p_{\text{atm}} + \rho g h

    • Pressure increases continuously with depth because the mass and weight of the liquid column above the point increase as depth increases (P_3 > P_2 > P_1).