# 16.1 Pressure Exerted by Gas Molecules When you inflate a tire, what is actually holding the car up? It is the trillions of tiny, fast-moving gas molecules inside, all crashing into the inner walls of the tire. This section explores how these countless tiny collisions create the force we measure as **pressure**. We will look at how the speed of these molecules relates directly to the temperature of the gas and derive the key formulas that connect the microscopic world of atoms to the everyday properties of gases we can see and feel. ### Origin of Gas Pressure Gas molecules are in continuous, random motion within a container. They collide with each other and with the walls of the container. These elastic collisions cause a change in momentum of the molecules, which, by Newton's second law, results in a force exerted on the walls. This force per unit area is defined as the pressure exerted by the gas. $$ \text{Pressure} = \frac{F}{A} = \left(\frac{\Delta p}{\Delta t}\right) \frac{1}{A} $$ The pressure depends on the number of molecules hitting the walls and the force of each collision. ### Derivation of Pressure Formula (Kinetic Theory) Consider an ideal gas of $N$ identical molecules, each of mass $m$, in a cubic box of side $l$ and volume $V = l^3$. **Collision of a single molecule:** A molecule moving towards a wall (face-1) with x-component velocity $v_{x1}$ has an initial momentum $mv_{x1}$. Upon perfectly elastic collision, its final momentum is $-mv_{x1}$. **Change in momentum of molecule:** $\Delta p = -mv_{x1} - (mv_{x1}) = -2mv_{x1}$. **Momentum imparted to the wall:** By conservation of momentum, the wall receives a momentum of $2mv_{x1}$. **Time between collisions:** The molecule travels a distance $2l$ (to the opposite wall and back) before colliding with the same wall again. The time taken is $\Delta t = \frac{2l}{v_{x1}}$. **Force by molecule on wall:** The rate of change of momentum for the wall due to one molecule is the force exerted by the molecule on the wall: $$ F_{x1} = \frac{2mv_{x1}}{2l/v_{x1}} = \frac{mv_{x1}^2}{l} $$ **Total force from N molecules:** Summing the forces from all $N$ molecules along the x-axis: $$ F_x = \frac{m}{l}(v_{x1}^2 + v_{x2}^2 + \ldots + v_{xN}^2) $$ **Mean Square Velocity:** Introducing the mean square velocity along the x-direction, $\langle v_x^2 \rangle = \frac{1}{N} \sum_{i=1}^N v_{xi}^2$. $$ F_x = \frac{mN}{l}\langle v_x^2 \rangle $$ **Isotropy of Motion:** For a large number of randomly moving molecules, the mean square velocities are equal in all three dimensions: $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$. **Total Mean Square Velocity:** The total mean square velocity $\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle = 3\langle v_x^2 \rangle$. So, $\langle v_x^2 \rangle = \frac{\langle v^2 \rangle}{3}$. **General Force:** Substituting this into the force equation: $$ F = \frac{mN}{l} \frac{\langle v^2 \rangle}{3} $$ **Pressure Equation:** Pressure $P = \frac{F}{\text{Area}} = \frac{F}{l^2}$. $$ P = \frac{mN}{l} \frac{\langle v^2 \rangle}{3} \left(\frac{1}{l^2}\right) = \frac{mN}{3l^3} \langle v^2 \rangle $$ Since $l^3 = V$ (volume of the container), the pressure exerted by an ideal gas is: $$ P = \frac{mN}{3V}\langle v^2 \rangle \tag{16.10} $$ This formula indicates that pressure is affected by volume: decreasing volume increases collision frequency, leading to higher pressure. ### Pressure in Terms of Average Translational Kinetic Energy Starting from $P = \frac{mN}{3V}\langle v^2 \rangle$, multiply and divide by 2: $$ P = \frac{2}{3} \frac{N}{V} \left\langle \frac{1}{2}mv^2 \right\rangle $$ Let $N_0 = N/V$ (number of molecules per unit volume). $$ P = \frac{2}{3} N_0 \left\langle \frac{1}{2}mv^2 \right\rangle $$ This shows that the pressure exerted by the gas molecules is directly proportional to the average translational kinetic energy $\left\langle \frac{1}{2}mv^2 \right\rangle$ of the gas molecules. $$ P \propto \langle \text{KE} \rangle $$ ### Pressure in Terms of Density Using $P = \frac{mN}{3V}\langle v^2 \rangle$ and recognizing that $\rho = \frac{mN}{V}$ (density of the gas): $$ P = \frac{1}{3}\rho \langle v^2 \rangle \tag{16.12} $$ ### Temperature in Terms of Average Translational Kinetic Energy **Ideal Gas Law:** $PV = nRT$, where $n = \frac{N}{N_A}$ (number of moles). Substitute $n$: $PV = N\left(\frac{R}{N_A}\right)T$. **Boltzmann Constant:** $k = \frac{R}{N_A} = 1.38 \times 10^{-23} \mathrm{JK}^{-1}$. Thus, $$ PV = NkT \tag{16.14} $$ From the pressure derivation, we also have: $$ PV = \frac{2N}{3} \left\langle \frac{1}{2}mv^2 \right\rangle \tag{16.15} $$ **Equating expressions for PV:** $$ NkT = \frac{2N}{3} \left\langle \frac{1}{2}mv^2 \right\rangle $$ Solving for T: $$ T = \frac{2}{3k} \left\langle \frac{1}{2}mv^2 \right\rangle \tag{16.16} $$ This fundamental relation states that the absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules: $$ T \propto \langle \text{KE} \rangle $$ ## Worked Examples **Example 1:** The density of air is $\rho = 1.296\ \mathrm{kgm^{-3}}$ at $0^\circ\mathrm{C}$. The root mean square velocity of air molecules is $484\ \mathrm{ms^{-1}}$. Determine the pressure of air. **Solution:** Given $\rho = 1.296\ \mathrm{kgm^{-3}}$ and $v_{rms} = 484\ \mathrm{ms^{-1}}$. Using the formula $P = \frac{1}{3}\rho v_{rms}^2$: $$ P = \frac{1}{3} \times 1.296 \times (484)^2 = 1.012 \times 10^5\ \mathrm{Pa} $$ **Example 2:** Calculate the average translational kinetic energy of a gas molecule at a temperature of $320\ \mathrm{K}$. **Solution:** Given $T = 320\ \mathrm{K}$ and Boltzmann constant $k = 1.38 \times 10^{-23}\ \mathrm{JK^{-1}}$. From the relation $T = \frac{2}{3k} \left\langle \frac{1}{2}mv^2 \right\rangle$, we can write: $$ \left\langle \frac{1}{2}mv^2 \right\rangle = \frac{3}{2}kT $$ $$ \left\langle \frac{1}{2}mv^2 \right\rangle = \frac{3}{2} \times (1.38 \times 10^{-23}\ \mathrm{JK^{-1}}) \times (320\ \mathrm{K}) $$ $$ \left\langle \frac{1}{2}mv^2 \right\rangle = 6.624 \times 10^{-21}\ \mathrm{J} $$ ## Formula Table | Concept | Formula | Description | | :------------------------- | :------------------------------------------------------------------------ | :------------------------------------------------------------------------------------------------ | | **Pressure (General)** | $P = \frac{F}{A} = \left(\frac{\Delta p}{\Delta t}\right) \frac{1}{A}$ | Force per unit area exerted by gas molecules due to momentum transfer. | | **Pressure (Kinetic)** | $P = \frac{mN}{3V}\langle v^2 \rangle$ | Pressure in terms of mass ($m$), number ($N$), volume ($V$), and mean square velocity. | | **Pressure (KE)** | $P = \frac{2}{3}N_0 \langle \frac{1}{2}mv^2 \rangle$ | Pressure in terms of number density ($N_0 = N/V$) and average translational kinetic energy. | | **Pressure (Density)** | $P = \frac{1}{3}\rho \langle v^2 \rangle$ | Pressure in terms of gas density ($\rho$) and mean square velocity. | | **Temperature (KE)** | $T = \frac{2}{3k} \langle \frac{1}{2}mv^2 \rangle$ | Absolute temperature in terms of Boltzmann constant ($k$) and average translational KE. | | **Boltzmann Constant** | $k = 1.38 \times 10^{-23}\ \mathrm{JK^{-1}}$ | Relates temperature to average kinetic energy at the molecular level. | These relationships are fundamental to understanding the macroscopic properties of gases from a microscopic, molecular perspective. See also for further discussion on molecular speeds.

16.1 Pressure Exerted By Gas Molecules | Statistical Mechanics And Thermodynamics | Physics 12