Pressure, its dependence on force, Pascal's law, pressure formulas, and the ideal gas law \( PV = nRT \)
1. Pressure and its dependence on force
Pressure \( P \) is the force per unit area acting perpendicular to a surface. It tells us how much force is distributed over a given area.
\[ P = \frac{F}{A} \]
where \( F \) is the magnitude of the force (in newtons, N) acting perpendicular to the surface, and \( A \) is the area (in m²). So pressure has units \( \text{N/m}^2 \), which is the pascal (Pa): \( 1\,\text{Pa} = 1\,\text{N/m}^2 \).
Dependence on force: For the same area, a larger perpendicular force gives a larger pressure. For the same force, a smaller area gives a larger pressure (e.g. sharp knife cuts because the force is concentrated on a small area). So \( P \propto F \) when \( A \) is fixed, and \( P \propto 1/A \) when \( F \) is fixed.
From force to pressure: If a fluid or solid pushes on a surface with a total force \( F \) spread over area \( A \), then the average pressure is \( P = F/A \). If the force is not uniform, we use \( P = dF/dA \) at a point (force per unit area on an infinitesimal area).
Example: A force of 500 N is applied on an area of 0.25 m². Then \( P = F/A = 500/0.25 = 2000\,\text{Pa} = 2\,\text{kPa} \). If the same force were applied on 0.01 m², \( P = 500/0.01 = 50\,000\,\text{Pa} = 50\,\text{kPa} \).
2. Pascal's law
Pascal's law (Pascal’s principle) says: in a fluid at rest that is incompressible (or effectively so, like water or oil), a change in pressure applied at one place is transmitted undiminished to every part of the fluid and to the walls of the container.
\[ \Delta P \text{ applied at one point } \Rightarrow \text{ same } \Delta P \text{ everywhere in the fluid} \]
So if you increase the pressure by \( \Delta P \) at the surface (e.g. by pushing a piston), that same \( \Delta P \) appears at every point in the fluid. This is the principle behind hydraulic lifts: a small force on a small piston creates a pressure increase that is transmitted to a large piston; the large piston has a larger area, so the total force on it is larger (\( F = P \times A \)).
Why it holds (idea): In a fluid at rest, the pressure at a point is the same in all directions (isotropy). If the fluid is incompressible, compressing it at one place would require the whole fluid to respond together; in equilibrium, the pressure difference that balances the applied force is the same throughout. So the applied \( \Delta P \) is transmitted.
Example — hydraulic lift: A small piston of area \( A_1 = 0.01\,\text{m}^2 \) is pushed with force \( F_1 = 100\,\text{N} \), so pressure increase \( \Delta P = F_1/A_1 = 10\,000\,\text{Pa} \). This same \( \Delta P \) acts on a large piston of area \( A_2 = 0.5\,\text{m}^2 \). Force on large piston \( F_2 = \Delta P \times A_2 = 10\,000 \times 0.5 = 5000\,\text{N} \). So a small force (100 N) lifts a large load (5000 N) — at the cost of moving the small piston a greater distance than the large one moves.
3. Connection between pressure, force, and Pascal's law
The link is:
\( P = F/A \): pressure is force per unit area. So \( F = P \times A \). For a given pressure, a larger area gives a larger total force.
Pascal's law: the pressure is the same throughout the fluid (after a change is applied). So at the small piston, \( P = F_1/A_1 \); at the large piston, the same \( P \) gives \( F_2 = P \times A_2 \). Hence \( F_2/A_2 = F_1/A_1 \), i.e. \( F_2 = F_1 \times (A_2/A_1) \). The force is multiplied by the ratio of areas.
So Pascal's law plus \( P = F/A \) explains why hydraulics can amplify force: the same pressure \( P \) applied to a larger area \( A_2 \) produces a proportionally larger force \( F_2 \).
4. Pressure formulas
Definition (all situations):
\[ P = \frac{F_\perp}{A} \]
Hydrostatic pressure (fluid at rest in a gravitational field): the pressure at depth \( h \) below the surface is greater than at the surface because of the weight of the fluid above:
\[ P = P_0 + \rho g h \]
where \( P_0 \) is the pressure at the surface (e.g. atmospheric), \( \rho \) is the density of the fluid (kg/m³), \( g \) is the gravitational field strength (about 9.8 N/kg), and \( h \) is the depth (m). So pressure increases linearly with depth.
Derivation of \( P = P_0 + \rho g h \): Consider a column of fluid of height \( h \) and cross-sectional area \( A \). Its weight is \( W = \rho V g = \rho A h g \). This weight is supported by the pressure difference between bottom and top: \( (P - P_0) A = W \), so \( P - P_0 = \rho g h \), hence \( P = P_0 + \rho g h \).
Atmospheric pressure at sea level is about \( 1.013 \times 10^5\,\text{Pa} = 101.3\,\text{kPa} \), often written as 1 atm. Gauge pressure is \( P - P_\text{atm} \); absolute pressure is \( P \).
Example: At what depth in water (\( \rho \approx 1000\,\text{kg/m}^3 \)) is the pressure twice atmospheric? \( P = P_0 + \rho g h = 2P_0 \Rightarrow \rho g h = P_0 \Rightarrow h = P_0/(\rho g) \approx 1.013\times 10^5/(1000 \times 9.8) \approx 10.3\,\text{m} \).
5. Ideal gas law: \( PV = nRT \)
For a dilute gas at low enough density and moderate temperature, pressure \( P \), volume \( V \), and temperature \( T \) are related by the ideal gas law:
\[ PV = nRT \]
where \( n \) is the number of moles of gas, and \( R \) is the universal gas constant: \( R \approx 8.314\,\text{J/(mol}\cdot\text{K)} \). Temperature \( T \) must be in kelvins (K).
So \( P \) depends on how much gas (n), how much space it has (V), and how hot it is (T). For a fixed amount of gas: \( P \propto T/V \); if \( T \) is constant, \( PV = \text{constant} \) (Boyle’s law); if \( V \) is constant, \( P \propto T \) (pressure law).
\[ \text{Also: } PV = N k_B T \quad \text{where } N = \text{number of molecules},\quad k_B = \frac{R}{N_A} \approx 1.38 \times 10^{-23}\,\text{J/K} \]
Here \( N_A \) is Avogadro’s number (\( \approx 6.022 \times 10^{23} \) per mole), so \( n = N/N_A \) and \( R = N_A k_B \).
Link to pressure as force per area: In kinetic theory, the pressure of a gas is due to molecules colliding with the walls. The force on the wall is the rate of change of momentum of the molecules; averaging over many collisions gives \( P \sim \rho \langle v^2 \rangle \), which can be shown to equal \( nRT/V \) for an ideal gas. So \( PV = nRT \) is consistent with \( P = F/A \) at the microscopic level.
Example: A container of volume 0.5 m³ holds 2 mol of an ideal gas at 300 K. \( P = nRT/V = 2 \times 8.314 \times 300 / 0.5 \approx 9986\,\text{Pa} \approx 10\,\text{kPa} \). If the same gas is compressed to 0.2 m³ at the same temperature, \( P = 2 \times 8.314 \times 300 / 0.2 \approx 24\,942\,\text{Pa} \approx 25\,\text{kPa} \).
6. Other useful relations
Density and moles: \( n = m/M \) where \( m \) is mass and \( M \) is molar mass. So \( \rho = m/V = n M/V \). From \( PV = nRT \), \( P = (n/V) RT = (\rho/M) RT \), so \( P = \rho R T / M \).
2. In a hydraulic system, the small piston has area 4 cm² and the large piston 100 cm². A force of 50 N is applied to the small piston. What force is produced on the large piston (assuming incompressible fluid)?
Solution: By Pascal's law, \( P \) is the same, so \( F_2/F_1 = A_2/A_1 = 100/4 = 25 \). So \( F_2 = 25 \times 50 = 1250\,\text{N} \).
3. What is the gauge pressure at a depth of 5 m in water? Use \( \rho = 1000\,\text{kg/m}^3 \), \( g = 10\,\text{m/s}^2 \).
Solution: Gauge pressure = \( P - P_\text{atm} = \rho g h = 1000 \times 10 \times 5 = 50\,000\,\text{Pa} = 50\,\text{kPa} \).
4. An ideal gas occupies 2 L at 100 kPa and 300 K. How many moles \( n \) are there? (Use \( R = 8.314\,\text{J/(mol}\cdot\text{K)} \); 2 L = 0.002 m³.)