Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above a given point on Earth's surface. It is a crucial aspect of atmospheric science and has various implications for weather, aviation, and everyday life. Atmospheric Pressure is about 1.01325 Pascal at sea level. Our body itself is quite used to this pressure but in very extreme essences like high altitudes, our nose might bleed because the pressure inside is higher than outside.

In this article, we have covered, the definition of Atmospheric Pressure, its important concepts and others in detail.

What is Atmospheric Pressure?

Air possesses weight and hence it exerts pressure like liquids because it possesses weight. The atmospheric pressure at a point is defined as the force acting normally on a unit area around that point, due to the total height of the air column of the atmosphere above it.

The value of it is maximum at sea level and decreases as we go to higher altitudes. Its symbol is Pa and the S.I. unit is newton per square meter (N m^-2) or pascal (Pa).

P_a = 1.013 × 10⁵ Nm^-2

Surface Pressure

Surface pressure is defined as the difference in interfacial tension between a clean interface and an interface in the presence of an emulsifier. The air pressure at a location on the surface (terrain and oceans) is called surface pressure, and it is proportional to the mass of air over that location.

Atmospheric models like general circulation models (GCMs) usually predict the non-dimensional logarithm of surface pressure, for numerical reasons.

On Earth, the average value of surface pressure is 985 hPa. The average pressure at mean sea level (MSL) within the international atmosphere (ISA) is 1013.25 hPa, or 1 atmosphere (atm), or 29.92 inches of mercury.

Pressure (p), mass (m), and the acceleration due to gravity (g), where A is the surface area are related by P = F/A = mg/A

Atmospheric pressure is thus proportional to the load per unit area of the atmospheric mass above that location.

What is Sea Level Pressure?

The air pressure is adequate to the pressure exerted by a mercury column of height “h”. Atmospheric pressure = h × d × g, where h is the height of the mercury column, d the density of mercury, and g the acceleration due to gravity.

Atmospheric Pressure = h.d.g = 0.76 × 13600 × 9.8 = 101293 Pa.

Standard sea-level pressure equals to 760 mm of mercury or 29.92 inches, 14.70 pounds per square inch, 1,013.25 × 103 dynes per square centimeter, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

Mean Sea Level Pressure

Mean sea-level pressure (MSLP) is that the air pressure at the mean water level (PMSL). Normally in weather reports on radio, television, and newspapers or the Internet we see this atmospheric pressure. When barometers within the home are set to match the local weather reports, they display pressure adjusted to water level, not the particular local air pressure.

The altimeter setting in aviation is an air pressure adjustment.

1013.25 mbar (101.325 kPa; 29.921 in Hg; 760.00 mmHg) average sea-level pressure. Weather reports (METAR), QNH is transmitted around the world in millibars or hectopascals (1 hectopascal = 1 millibar), except within the US, Canada, and Colombia where it's reported in inches of mercury (to two decimal places), in aviation. The US and Canada also report sea-level pressure SLP, which is adjusted to water level by a special method, within the remarks section, not within the internationally transmitted part of the code, in hectopascals or millibars. However, in Canada's public weather reports, water level pressure is instead reported in kilopascals.

In the US weather code remarks, three digits are all that are transmitted; decimal points and therefore the one or two most vital digits are omitted: 1013.2 mbar (101.32 kPa) is transmitted as 132; 1000.0 mbar (100.00 kPa) is transmitted as 000; 998.7 mbar is transmitted as 987; etc. the very best sea-level pressure on Earth occurs in Siberia, with record highs on the brink of 1085 mbar (108.5 kPa; 32.0 in Hg), where the Siberian High often attains a sea-level pressure above 1050 mbar (105 kPa; 31 in Hg). rock bottom measurable sea-level pressure is found at the centers of tropical cyclones and tornadoes, with a record low of 870 mbar (87 kPa; 26 in Hg).

Variation Atmospheric Pressure in Altitude

Pressure on Earth varies with the altitude of the surface, so air pressure on mountains is usually lower than the air pressure on ground level. Pressure varies smoothly from the surface to the very best of the mesosphere. Although the pressure changes with the weather, NASA has averaged the conditions for all parts of the world year-round. As altitude increases, air pressure decreases. At a given altitude one can calculate the atmospheric pressure. Temperature and humidity also affect the air pressure, and it's necessary to understand these to compute an accurate figure. The graph at right was developed for a temperature of 15 °C and a humidity of 0%.

The pressure decreases by about 1.2 kPa (12 hPa) for every 100 meters, at low altitudes above water level. For higher altitudes within the troposphere, the next equation (the barometric formula) relates atmospheric pressure p to altitude h:

p = p_{0} \times (1 - \frac{L\times h}{T_{0}})^{\frac{g\times M}{R_{0}\times L}}

= p_{0} \times (1- \frac{g\times h}{c_{p}\times T_{0}})^{\frac{c_{p}\times M}{R_{0}}}

\approx p_{0}\times exp\times (-\frac{g\times h\times M}{T_{0}\times R_{0}})

What is Barometer?

The earth’s air pressure at an associate degree given place is measured with an instrument known as the measuring instrument. The barometer is a perfect and commonplace instrument to measure air pressure. Within the barometer, the modification within the pressure is indicated on a dial by a pointer. The barometer was fictitious by physicists around 1644.

A simple measuring instrument consists of a glass tube of length 100 cm with one finish closed, and a trough with mercury. The tube is stuffed with pure and dry mercury. Its open finish is closed with a thumb and inverted vertically within the trough stuffed with mercury.

It is ascertained that a number of the mercury from the glass tube flows down into the trough and a set length of mercury column remains within the tube. The empty area on top of the mercury column is named the Torricellian vacuum.

Consider 2 points, “A” and “B” on the surface of the mercury. Note that each point square measures at an identical level. There's the air on top of purpose “A” and also the mercury column on top of purpose “B”. PA is that the pressure exerted by the atmosphere at purpose A and the metallic element is that the pressure exerted by the peak of the mercury column within the tube at purpose B. Since “A” and “B” square measure on an identical horizontal plane,

P_A = P_B.

Therefore, the air pressure is expressed in terms of the length of the mercury column. The air pressure level is 76 cm of mercury. The standard air pressure on the surface varies between 940 and 1040 hPa (mbar). The standard air pressure at ground level is 1013 hPa (mbar).

Applications on Atmospheric Pressure

Here are the applications of barometric pressure:

• Meteorology: These observations of wind composed since the late 19th century are common for weather prediction, especially when combined with winds observations.

• Weather Maps: The most important thing barometric indicators give is that they help generate an isobar (lines of equal pressure) maps showing regions of strong or weak pressure.

• Weather Forecasts: Extremely fast alterations in the weather, by 3.5 hPa or more, normally indicates either the oncoming low regime or good after a cold front.

• Mining Safety: It is the case that the changing in atmospheric pressure may lead to a number of gas concentration, notably the firedamp. To ensure miners’ safety, pressure in the mine is always monitored to secure the task of pre-emptive action.

• Skin Diving: Smartphones with the Android systems are used to transmeter pressure differences during scuba diving, the activities comprising of air tanks monitoring and hydrostatic pressure measurement as an indicator of depth.

Equation of Air Pressure

When air pressure is measured by a barometer, the pressure is additionally mentioned because of the "barometric pressure". Assume a barometer with a cross-sectional area A, a height h, full of mercury from the lowest at Point B to the very best at Point C. The pressure at the rock bottom of the barometer, Point B, is capable of the air pressure. The pressure at the very top, Point C, is often taken as zero because there's only mercury vapor above now and its pressure is extremely low relative to the atmospheric pressure. Therefore, one can find the air pressure using the barometer and this equation:

P_atm = ρgh

Where the density of mercury is ρ, the gravitational acceleration is g, and the height of the mercury column above the free surface area is h. The physical dimensions (length of the tube and cross-sectional area of the tube) of the barometer itself do not affect the height of the fluid column in the tube.

In thermodynamic calculations, a commonly used unit of measurement is the "standard atmosphere". At 0 °C this is the pressure resulting from a column of mercury of 760 mm in height. Use ρHg = 13,595 kg/m3 for the density of mercury and use g = 9.807 m/s² for gravitational acceleration.

If water were used (instead of mercury) to satisfy the quality air pressure, a water column of roughly 10.3 m (33.8 ft) would be needed.

Standard atmospheric pressure as a function of elevation:

Note: 1 torr = 133.3 Pa = 0.03937 in Hg

Read More:

• How to calculate Atmospheric Pressure?
• Difference between Absolute Pressure and Atmospheric Pressure
• Atmosphere

Examples on Atmospheric Pressure

Example 1. At a given place, a mercury barometer records a pressure of 0.50 m of Hg. If mercury in the barometer is replaced by water What would be the height of the water column? Take R.D. of mercury = 13.6.

Solution:

Relative Density of Hg = 1.36 = 13.6 × 10³ kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Height of mercury column = 0.50 m

Pressure, P = ρgh

P = {0.5×(1.36×10³)×9.8} pascal

P = 6.664 × 10³

Let h be the height of the water column

Then, P = h x density of water x g

6.664 × 10³ = h × 10³ × 9.8

h = 0.68 m

Example 2. The atmospheric pressure is 76 cm of Hg at sea level. What is the height of a hill where the barometer reads 60 cm Hg if air pressure falls by 10 mm of Hg per 120m of ascent? State the assumption made by you.

Solution:

Atmospheric pressure, P = 76 cm Hg

Pressure falling rate = 10 mm of hg per 120 m of ascent = 1 cm of Hg per 120 m of ascent

Pressure at hill, P` = 60 cm Hg

Total fall in pressure = P – P` = (76-60) cm Hg = 16 cm Hg

Now, the fall in pressure is 1 cm Hg for every 120 m increases in height.

Thus, if the fall in pressure is 6 cm Hg increase in height shall be (16 × 120) m = 1920 m

Example 3. The atmospheric pressure is 1.04 × 10⁵ Pa at sea level. Assuming the density of air to be unvarying and alike to 1.6 kg m^-3 and g = 10 m s^-2, find the height of the atmosphere.

Solution:

Atmospheric pressure, P = 1.04 × 10⁵ Pa

g = 10 m s^-2

Density, ρ = 1.6 kg m^-3

Let h be the height of the atmosphere.

P = ρgh

\therefore h = \frac{P}{\rho \times g} = \frac{1.04\times 10^{5}}{1.6\times 10}     = 6500 m

Example 4. At a height of 108 m above the sea level assuming the density of air to be 1.36 kg m^-3 find the fall in barometric height in mm of Hg. Take density of Hg = 13.6 × 10³ kg m^-3.

Solution:

Let h = 108 m be the height above sea level

P_h – P_sea = ρ_air gh

ρ_mgh_f - ρ_mgh_i = ρ_airgh

ρ_mg × Δh = ρ_airgh

 \Delta h = \frac{\rho _{air}\times h}{\rho _{m}} = \frac{1.36\times 10^8}{13.6\times 10^{6}}      = 10.8 x 10^-6 mm of Hg

Example 5. Convert 2 mm of Hg into pascal. Take the density of Mercury, Hg = 13.6 × 103 kg m^-3 and gravitation, g = 9.8 m s^-2.

Solution:

Density of Mercury, Hg = 13.6 × 103 kg m^-3

Acceleration due gravity, g = 9.8 m s^-2

Height of the mercury column = 2mm = 0.002 m

Therefore, Pressure, P = hρg

P = 0.002 × (13.6 × 10³) × 9.8 pascal

P = 254.8 Pa

Example 6. Solve the following:

(a) Calculate the height of a column of water which will exert on its base the same pressure as the 72.5cm column of Hg. The density of Hg is 13.6g cm^-3.

(b) If the cross-section of the water column is made wider will the height of the water column in the previous part change?

Solution:

(a) We know that the pressure exerted by the water column, P = hρg

Density of water = 1

As the pressure of H₂O and Hg is the same,

h_w ρ_w g = h_m ρ_m g

h_w × 1 × g = h_m ρ_m g

h_w = h_m ρ_m

= \frac{72.5}{100}\times 13.6

= 9.86

Height of the water column = 9.86m

(b) Height of the water column will be unaffected if the water column is made wider.

Example 7. The water pressure on the bottom floor is 60,800 Pa and on the primary floor is 15,800 Pa. Find the height of the first floor. (Take: density of H₂O = 1000kg m^-3, g = 10m/s²)

Solution:

To find the height of the first floor:

Given: P_w on ground floor P_wg = 60,800 Pa

P_w on first-floor P_wf = 15,800Pa

From the formula of pressure:

P = hρg

In order to know the height of the first flowe have to calculate the difference in pressure

P = P_wg - P_wf

= 60800 – 15800

= 45000Pa

Substituting this value in the formula for the pressure to calculate height:

P = hρg

45000 = 1000 × 10 × h

h = 4.5m

Height of the floor is 4.5 m.