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Pressure refers to the force applied per unit area inside a particular fluid or the perpendicular force per unit area. Whereas density implies mass by volume of any particular material. The SI unit for pressure is Pascal and the SI unit for density is kilogram per cubic meter. Pressure and density are directly related to each other. The change of one particular quantity will create a difference in the other too.
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Key Takeaways: Pressure, Density, Area, Force, Fluid
Pressure
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Pressure of a fluid implies force applied per unit area; the force is said to act perpendicular to the surface. Pressure is a scalar quantity. It can be represented by:
P = F/A
(force per unit area), where F is the perpendicular force applied on the surface and A is area of the surface.
The SI unit is N/m2 . Another basic unit to use is Pascal, Pa. (1 Pa = 1 N/m^2)
It changes based on the following factors:
- Pressure exerted increases upon an increase in the perpendicular force applied and decreases in reduction of the force
- Pressure exerted rises upon reduction in the surface area and falls upon expansion of area of force applied.
- When fluid (gas) amount is increased in a container, the pressure increases too against the walls of the container and reduction in amount or weight of the substance decreases the pressure against the walls.
Density
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Density of a substance (solid or fluid) implies mass by volume, or mass per unit volume, of that substance. It is a scalar quantity and can be represented by ‘\(\rho\)’ in the following way:
\(\rho\)= M/V
(mass per unit volume), where \(\rho\) is density, M is mass and V is volume occupied by the given solid or fluid. The SI unit is kg/m3, and sometimes by g/m3.
Density changes depending upon the following factors:-
- Density increases upon increase in mass of substance and decreases upon decrease in mass of the given substance.
- Density increases upon decrease in volume while it decreases when the volume of the substance rises.
Difference between Pressure and Density
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Pressure and Density have a direct relation; change in pressure of the fluid induces a change in density of the fluid too. Similarly, change in density of the fluid also causes a change in the pressure of the fluid.
You can understand in this way:
- Increase in Pressure → Increase in Density
- Increase in Density → Increase in Pressure
- Decrease in Pressure → Decrease in Density
- Decrease in Density → Decrease in Pressure
A direct relation between pressure and density results in a linear graph:

Relation between Pressure and Density
This direct relation between pressure and density is established with the help of Boyle’s law. Boyle's law states that the pressure induced by a given mass of gas is inversely proportional to its volume, provided that the temperature is kept constant.
Therefore, the graph will be nonlinear when plotted between Pressure and Volume of the gas:
In other words, the pressure and volume of the gas continue to share an inversely proportional relation till the time the mass of the gas and the temperature are kept constant.
For example: When a balloon, filled with air, is compressed, we see that the size of the balloon begins to distort because of the increase in pressure. And, the distortion of the balloon states that the volume of the balloon (due to air present inside it) reduces. Thus, higher the pressure, lesser is the volume.
Relation between Pressure and Density through Boyle's Law
We know that, at a given mass and constant temperature condition, Pressure is inversely professional to volume. Then,
P ∝ 1/V
PV ∝ 1
(removing the proportionality sign, we will get)
PV = C (where, P is pressure, V is volume and C is constant)
We, hence, get that P times V (product of Pressure and Volume) will always be constant (thus, the product will be a constant too).
This is known as Boyle’s Law and the product of PV is called Boyle’s constant.
Now, we know that,
P ∝ 1/V
So, multiply mass (M) on both sides. We will get,
P ∝ M/V
And, we know that mass by volume is density, represented by ‘\(\rho\)’ .
Thus,
P ∝ \(\rho\)
We have proved that pressure is directly proportional to density.
Relation between Pressure and Density using Fluid Mechanics
Fluid mechanics is studying the forces and flow of the fluids (essentially liquids, here). For pressure, we know that there is Pressure variation as per the depth of the liquid. The deeper you go in a liquid, the higher the pressure gets.
The pressure exerted as a result of the weight of the liquid, in a constant density condition, and depending upon the depth of the liquid, is given by the expression of:
P = ρgh
(where, P is pressure exerted, ρ is the density of the liquid, h is depth of the liquid and g is the acceleration due to gravity)
Here, we imply that pressure is again directly proportional to the density of the liquid, if we do not change the value of ‘g’ and ‘h’ any further (constant).
Another way to provide relation between pressure and density is through the following:
P = F/A (where P is pressure, F is perpendicular force applied and A is the surface area)
F = Mg (where M is mass and g is acceleration due to gravity)
ρ = M/V (where ρ is density and V is volume)
So, M = ρV and, F/g = M
Then, ρV = Fg and ρVg = F
So, F = PA
Then, PA = ρVg
We know that, A = L x L (where L is length) and V = L x L x L
Then,
P (L x L) = ρ (L x L x L)g
P = ρLg
Putting L = h, where h is height, we ultimately get,
P = ρgh
Hence, proved.
Things to Remember
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- Pressure implies force acted on a surface per unit area while density states mass by volume of a solid or fluid.
- Pressure and density showcase a linear graph, stating that they are directly proportional to each other. Thus, a change in one quantity induces a change in another as well.
- The relation between pressure and density is established by Boyle’s law, which states that at a constant temperature for a given mass, the pressure remains inversely proportional to the volume.
- The similar relation can be established with pressure varying with depth and the density of the liquid.
Also Read:
Sample Questions
Ques. A 50 kg girl has donned heels. She tries to balance herself on one heel which has a circular diameter (1 cm). Calculate the pressure exerted. (3 marks)
Therefore Force on the heels, F = mg = 50*9.8 = 490 N
Diameter, D = 1.0 cm = 1*10-² m
Area, A = (\(\pi\)D²)/4 = 3.14*(1*10-²)²
A = 7.85*10-2m²
Pressure, P = F/A = 490/(7.85*10-2) = 6.24*106 Pa
Ques. Mercury is utilised in the Torricelli's barometer which was duplicated by Pascal using French wine (density = 984 kg/m^3). Find the height of the wine column. (Atmospheric pressure = 1.01 x 10^5 Pa). (3 marks)

Ques. State and explain the relation between pressure and density using Boyle’s law. (4 marks)
At a given mass and constant temperature condition, Pressure is inversely professional to volume. Then,
P ∝ 1/V
PV ∝ 1
PV = C (where, P is pressure, V is volume and C is constant)
Hence, Boyle’s law is proved.
Now,
P ∝ 1/V
So, multiply mass (M) on both sides. We will get,
P ∝ M/V
And, we know that mass by volume is density, represented by ‘\(\rho\)’ .
Thus,
P ∝ \(\rho\)
We have proved that pressure is directly proportional to density.
Ques. The sea-level atm density is 1.29 kg/m^3. If the altitude does not change, find the height of the atmosphere. (2 marks)
g is referred to as the acceleration due to gravity
Given,
\(\rho\) = 1.29 kg/m³
P = 1.013 * 105Nm-² and g = 9.8m/s²
P = \(\rho\)gh = (1.013*105)/(1.29*9.8) = 8013 km
Therefore the height of the atmosphere is 8013 km above sea level.
Ques. A hydraulic lift can lift the weight of cars upto 3000 kg mass. The piston that carries the load has an area of cross-section of 425 cm^2. Calculate the pressure the smaller piston exerts. (2 marks)
P = (3000*9.8)/(425*10-4)Nm-²
= 6.92*105Pa
Ques. The femurs have a cross-sectional area of 10 cm^2 each to support a body of 40 kg mass. Find the pressure exerted. (2 marks)
P = F/A = mg/A
P = (40/10)/(2*10*10-4)
P = 2*105 Nm²
Ques. Explain the variation of pressure with depth. (5 marks)
F = Mg (where M is mass and g is acceleration due to gravity)
ρ = M/V (where ρ is density and V is volume)
So, M = ρV and, F/g = M
Then, ρV = Fg and ρVg = F
So, F = PA
Then, PA = ρVg
We know that, A = L x L (where L is length) and V = L x L x L
Then,
P (L x L) = ρ (L x L x L)g
P = ρLg
Putting L = h, where h is height, we ultimately get,
P = ρgh
Hence, proved. The P is proportional to height. The more the depth, the more pressure. Alternatively, Pa + ρgh = P, where ρa is constant atmospheric pressure at the surface (1.01 x 10^5 pa).
ρgh is actually gauge pressure that results from P-Pa. In this case too, the more the height, the more pressure.






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