**Pressure**

In its simplest definition, pressure is the force per unit area. The force is acting perpendicular to the given area. Mathematically, pressure can be expressed as

\(P = {F \over A}\)

where ** F** is the force measured in

**newtons (N)**,

**is the area measured in**

*A***square meters (m**, and

^{2})**is the pressure which is measured in**

*P***pascals (Pa)**. The unit Pascal is named after Blaise Pascal who is the pioneer in studying pressure.

Pressure is important in understanding the behavior of fluids when at rest. Here are some of the properties of pressure of fluids:

1. The pressure of the fluid is equal in any material of equal depths. It depends only on the depth and the density of the fluid in a material or container. This implies that, as the depth of the fluid increases, the pressure also increases.

2. The pressure exerted by the fluid is equal in all directions in a material.

3. The pressure exerted is perpendicular to any surface the fluid is in contact with.

**Example.**

An aquarium with depth of 1.0 m and cross-sectional area of 8 m^{2}, is filled with water. How much pressure is exerted by the water on the bottom of the tank?

Solution:

- We first solve for the force using the equation F = mg. Since the value of mass is not given, we also use the formulas for mass in terms of the density which is m = \(\rho\)V. Thus, force can be calculated as \(F = mg = (\rho V)g\). The density of water is 1.0 x 10
^{3}kg/m^{3}.

\( F = (1.0 x 10^3\; kg/m^3)(8\;m^3)(9.8\;N/kg)= 7.84 \times 10^4\;N \)

- Use the equation for pressure, \(P = {F \over A}\).

\(P = {7.84 \times 10^4\;N \over 8\;m^2} = 9.8 \times 10^3\;N/m^2\; \text{or} \;9.8 \times 10^3\;Pa\)

**Atmospheric Pressure**

The atmosphere is composed of different mixture of gases which is also known as air. It is a fluid which exerts pressure. The pressure exerted by the weight of the atmosphere is called the *atmospheric pressure.*

Since the value of g varies as the height of the atmosphere, atmospheric pressure cannot be calculated using the equation for pressure in fluid at specific depth which is expressed as \(p=\rho gd\). However, a device called ** barometer** can be used to measure it. It is a thin, strong-walled glass tube, sealed at one end and open at the other end. It is then filled with liquid and is carefully put inverted on a container which also contains the same liquid as that in the barometer. Liquid mercury is usually used in barometers due to its high density.

Atmospheric pressure differs with different altitude but through various measurements made, it has been determined that the average atmospheric pressure at sea level holds a 760 mm high column of mercury. The unit often used in measuring the atmospheric pressure is the atmosphere, abbreviated as atm, which is equal to the height of mercury column it could support.

\(1\; atm = 760\; mm\; Hg\)

In pascals, 1 atm can be converted using \(1\;atm = \rho g d\).

where \(\mathbf{\rho}\) is the density of mercury= 13.6 \(\times\) 10^{3} kg/m^{3} and d is the height of mercury column in 1 atm and is equal to 0.760 m, and g = 9.8 N/kg. Substituting all these values to the equation, we have

\(1\;atm = (13.6 \times 10^3 kg/m^3)( 9.8 N/kg)( 0.760 m)= 1.013 \times 10^5\;N/m^2,\;\; \text{or}\;\; 1.013 \times 10^5\;Pa\)

**Absolute Pressure vs. Gauge Pressure**

In dealing with pressure in fluids, one must be able to distinguish the difference between absolute pressure and gauge pressure. **Absolute pressure** is the total pressure, including atmospheric pressure, exerted to all the objects on the Earth’s surface. It is the sum of the atmospheric pressure and the gauge pressure.

\(p_{\text{absolute}}=p_{\text{atmospheric}} + p_{\text{gauge}}\)

**Gauge pressure** is any additional pressure relative to the atmospheric pressure. It is the difference between absolute pressure and atmospheric pressure. Negative signs are usually omitted.

\(p_{\text{gauge}} = p_{\text{absolute}} - p_{\text{atmospheric}}\)

It can also be solved using the equation: \(p = \rho gd\).

**Example.**

The swimming pool is 1.5 m deep. Calculate the:

- Gauge pressure
- Absolute pressure of the fluid.

Solution:

- The gauge pressure can be solved using the equation \(p = \rho gd\), where \(\rho\)=1.0 x 10
^{3}kg/m^{3}, g = 9.8 N/kg, and d = 1.5 m.

\( p = (1.0 \times 10^3 kg/m^3)( 9.8 N/kg)(1.5 m)=1.47 \times 10^4\;Pa\)

- To solve for the absolute pressure, we take the sum of the atmospheric pressure and gauge pressure. The pressure that is exerted at the surface of the pool is normal atmospheric pressure and is equal to 1.0 x 10
^{5}Pa.

\(p_{absolute}=p_{atmospheric} + p_{gauge}\\ \quad\quad\quad\;=1.0 x 10^5 Pa + 1.47 \times 10^4\;Pa\\ \quad\quad\quad\;=\mathbf{1.147 \times 10^5\;Pa}\)

In case calculator is not allowed, to get the sum of two scientific notation with different exponents, first convert all the numbers to the same power of 10.

\(p_{\text{absolute}}=10 \times 10^4 Pa + 1.47 \times 10^4\;Pa\)

Add only the digit terms and copy the exponential term.

\(p_{\text{absolute}}=11.47 \times 10^4\;Pa\)

In this case, more than one non-digit number is on the left of the decimal point. Move the point to the left, so that there’s only one nonzero digit left on the left of the point. Add the number of moves to the exponent. Thus, the answer will become

\(1.147 \times 10^5\;Pa\).