Pressure Calculation Example

In Physics, pressure is the force exerted on a certain Area. The most common case of Pressure is the Weight of a body on the surface it occupies on the planet.

Pressure can be exerted by matter in the three physical states: solid, liquid and gas.

Pressure Manifestation

Pressure can occur in very different circumstances:

• In one column, there may be different liquids that do not mix, resting one on another. Each liquid will exert pressure on the one below it. Whoever is at the bottom, will receive the joint pressure of all those above.
• In a closed container, such as a balloon, there may be a gas or gas mixture that is going to exert a pressure on its walls.
• In an internal combustion engine, the descending piston generates a pressure on the gasoline-air mixture. As the spark enters the system and explodes, the chemical reaction will put pressure on the piston, lifting it up again.
• All the gases present in the atmosphere generate a pressure on the surface of the Earth. This pressure is called Barometric Pressure or Atmospheric Pressure.

Barometric or Atmospheric Pressure

The actual pressure of the atmosphere is measured with an instrument called Barometer, devised by E. Torricelli in 1644. The scientist made this instrument using a 1 meter long tube, sealed on one side. He filled the tube with Mercury, and dipped the open side into a vat full of more Mercury.

The Mercury in the tube descended by gravity, until it adjusted to a level of 760 millimeters. The pressure of the atmosphere subdued the Mercury in the Cuba, pushing it until the tube was adjusted to that height. Since then it has been established that the Standard Atmospheric Pressure has a value of 760mmHg.

The Barometric or Atmospheric Pressure is measured with the Barometer instrument, or also with the so-called Barograph, which in addition to the Pressure measurement includes a pen with ink to trace the value of Atmospheric Pressure on a graph during the course of the weather.

Gauge pressure

The Gauge Pressure is that which is exerted on the walls of a closed container. Generally it refers to that exerted by gases, since they have the property of covering the entire volume of the container that contains them.

Depending on the mass of gas contained, it will be the amount of gas particles that apply force on the walls of the container, and therefore the magnitude of the Gauge Pressure to be measured.

The gas can be in a state of rest in a tank, or in motion, constantly moving along a pipe system.

Gauge Pressure is measured with devices called Gauges, which are circular like a clock, and have on the dial the scale in the units in which Pressure is measured. The Pressure Gauge responds to the thrust of the fluid or gas and returns a reading with its indicator needle.

Pressure Measurement Units

Millimeter of Mercury (mmHg): It was the first unit for Barometric Pressure thanks to the design of the Torricelli Barometer. Standard Barometric Pressure corresponds to 760mmHg.

Pascal (Pa): It is the unit established for Pressure in general, according to the international System of Units. According to his concept of "Force over Area", it is equivalent to 1 Newton over square meter (1 Pa = 1 N / m²). The equivalence in Pascals of the Atmospheric Pressure is 101,325.00 Pascals.

Pounds on Square Inch (lb / in², psi): It is the unit in the English System of Units for pressure. It is the most used to calibrate industrial pressure gauges and devices for conventional use. It is called “psi” from its English terms: “pounds square inches”. The equivalence in psi of atmospheric pressure is 14.69 lb / in².

Bars (bar): Bar is an alternative unit for measuring pressure. It is used in Literature to refer to large magnitudes of pressures, so as not to use such large numbers. The bar equivalent of atmospheric pressure is 1,013 bar.

Atmospheres (atm): It is the unit established for atmospheric pressure, located exactly at the Barometric pressure measured in the area in which the calculations are made. Its value is always set as 1 atm, and has different equivalences with other units. Of course, if the atmospheric pressure is measured in other units, the numerical data will be different.

Pressure calculations

Pressure will be calculated differently, depending on the physical state of the substance that exerts it: solid, liquid or gaseous. Of course, the formulas can be used for all cases, but to be better explained, we resort to classifying the calculations like this.

Pressure exerted by solids:

For solids, the formula is used

P = F / A

Define pressure as a Force exerted on an Area. Solids naturally encompass a defined area, so the force to be exerted will be their Weight, unless an additional force is also acting on the solid.

To obtain the pressure in Pascals (Pa = N / m²), it is necessary that the Force be in Newton (N) and the Area in square meters (m²).

Pressure exerted by liquids:

For liquids, the formula is used

P = ρ * g * h

Define pressure as the product of Density, the force of Gravity and the Height that the liquid covers in the column where it is confined. If there are two or more liquids in the column, separated by densities, the formula works for each liquid by its side.

So that the pressure is obtained in Pascals (Pa = N / m²), it is necessary that the Density be in Kilograms over cubic meter (Kg / m³), gravity in meters over second squared (m / s²) and the height in meters (m).

Pressure exerted by gases:

The pressure of a gas, if it behaves like an ideal gas, can be calculated with the expression of the ideal gas:

PV = nRT

Having the data of number of moles of gas, Temperature and Occupied volume, it can be calculated immediately. If it is a Real Gas, it will be necessary to resort to the equations for Real Gas, which are more complex than the simple ideal gas relation.

For Pressure to be in Pascals, Volume must be in cubic meters (m³), the Temperature in absolute degrees Kelvin (K), and the ideal gas constant must be R = 8.314 J / mol * K.

Examples of how to calculate pressure

There is a solid body with a weight of 120 N, and it covers a surface area of ​​0.5 m². Calculate the pressure exerted on the ground.

P = F / A

P = (120 N) / (0.5 m²) = 240 N / m² = 240 Pa

There is a solid body with a weight of 200 N, and it covers a surface area of ​​0.75 m². Calculate the pressure exerted on the ground.

P = F / A

P = (200 N) / (0.75 m²) = 266.67 N / m² = 266.67 Pa

It has a solid body with a weight of 180 N, and covers a surface area of ​​0.68 m². Calculate the pressure exerted on the ground.

P = F / A

P = (180 N) / (0.68 m²) = 264.71 N / m² = 264.71 Pa

It has a solid body with a weight of 230 N, and covers a surface area of ​​1.5 m². Calculate the pressure exerted on the ground.

P = F / A

P = (230 N) / (1.5 m²) = 153.33 N / m² = 153.33 Pa

There is a column with two liquids, with densities of 1000 Kg / m³ and 850 Kg / m³. The liquids gather heights of 0.30 m and 0.25 m respectively. Calculate the Pressure at the bottom of the container.

P = (ρ * g * h)₁ + (ρ * g * h)₂

P = (1000 Kg / m³) * (9.81 m / s²) * (0.30 m) + (850 Kg / m³) * (9.81 m / s²) * (0.25 m)

P = 2943 Pa + 2085 Pa = 5028 Pa

There is a column with two liquids, with densities of 790 Kg / m³ and 830 Kg / m³. The liquids gather heights of 0.28 m and 0.13 m respectively. Calculate the Pressure at the bottom of the container.

P = (ρ * g * h)₁ + (ρ * g * h)₂

P = (790 Kg / m³) * (9.81 m / s²) * (0.28 m) + (830 Kg / m³) * (9.81 m / s²) * (0.13 m)

P = 2170 Pa + 1060 Pa = 3230 Pa

There is a column with two liquids, with densities of 960 Kg / m³ and 750 Kg / m³. The liquids gather heights of 0.42 m and 0.20 m respectively. Calculate the Pressure at the bottom of the container.

P = (ρ * g * h)₁ + (ρ * g * h)₂

P = (960 Kg / m³) * (9.81 m / s²) * (0.42 m) + (750 Kg / m³) * (9.81 m / s²) * (0.20 m)

P = 3960 Pa + 1470 Pa = 5820 Pa

There is a column with two liquids, with densities of 720 Kg / m³ and 920 Kg / m³. The liquids gather heights of 0.18 m and 0.26 m respectively. Calculate the Pressure at the bottom of the container.

P = (ρ * g * h)₁ + (ρ * g * h)₂

P = (720 Kg / m³) * (9.81 m / s²) * (0.18 m) + (920 Kg / m³) * (9.81 m / s²) * (0.26 m)

P = 1270 Pa + 2350 Pa = 3620 Pa

There are 14 moles of an ideal gas, covering a Volume of 2 m³ at a Temperature of 300 K. Calculate the pressure exerted against the walls of the container.

PV = nRT P = (nRT / V)

P = (14 mol) (8.314 J / mol * K) (300 K) / 2 m³ = 17459.4 Pa

There are 8 moles of an ideal gas, covering a Volume of 0.5 m³ at a temperature of 330 K. Calculate the pressure exerted against the walls of the container.

PV = nRT P = (nRT / V)

P = (8 mol) (8.314 J / mol * K) (330 K) / 0.5 m³ = 43897.92 Pa

There are 26 moles of an ideal gas, covering a Volume of 1.3 m³ at a temperature of 400 K. Calculate the pressure exerted against the walls of the container.

PV = nRT P = (nRT / V)

P = (26 mol) (8.314 J / mol * K) (400 K) / 1.3 m³ = 66512 Pa

There are 20 moles of an ideal gas, encompassing a Volume of 0.3 m³ at a Temperature of 350 K. Calculate the pressure exerted against the walls of the container.

PV = nRT P = (nRT / V)

P = (20 mol) (8.314 J / mol * K) (350 K) / 0.3 m³ = 193993.33 Pa