13 3 Pressure In A Gas – Class 12TH Physics

13.3 Pressure in a Gas and Derivation of $PV = \tfrac{1}{3}Nm\langle v^{2}\rangle$

We can use the kinetic theory of the gas to derive an equation which relates the macroscopic properties of a gas (pressure and volume) to the microscopic properties of its molecules (mass and speed). It is assumed that a single molecule is moving in a cube-like box of each side $L$ (Fig. 13.2). This molecule has mass $m$ , and is moving with speed " $v$ ". It moves back and forth, colliding at regular intervals with the walls and therefore, it contributes to the pressure of the gas. We are going to work out the pressure exerted by this one molecule on the wall of the box and then deduce the total pressure produced by all the molecules.

According to the kinetic theory of gases, a gas consists of a large number of tiny particles (atoms or molecules) that are in constant random motion. These particles move freely in all directions and frequently collide with one another and with the walls of the container. When gas molecules strike the walls, they exert a force on them due to the change in momentum during each collision. The pressure of a gas is defined as the force exerted per unit area of the container walls as a result of these collisions. Since the collisions are elastic, no energy is lost, and the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas i.e., $\langle\mathrm{K.E.}\rangle \propto T$ . Thus, an increase in temperature increases the speed of the molecules, leading to more collisions, and consequently, a higher pressure.

The ideal gas obeys the gas law which would relate the gas temperature with other parameters and the gas law is:

PV = nRT \quad \ldots (13.1)

where $P$ is the pressure, $n$ is number of moles, $R$ is general gas constant equals to $N_{A}k_{B}$ , $k_{B} = 1.38 \times 10^{-23}\,\mathrm{J\,K^{-1}}$ is the Boltzmann constant, ( $N_{A} = 6.02 \times 10^{23}$ molecules) is the Avogadro number and $T$ is the temperature of the gas. Therefore,

PV = nN_{A}k_{B}T

The number of moles in terms of number of particle $N$ is defined as; $n = \dfrac{N}{N_{A}}$ , then

PV = Nk_{B}T \quad \ldots (13.2)

As pressure is force per unit area, exerted on walls of the container, therefore, each time a gas particle (molecule or atom) collides with walls of the container, and exerts a force on the walls of the container. This force spreads over whole wall area of the container.

Brain teaser

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What is the difference between mole (gram-molecule) and Avogadro's number? Does Avogadro's number depend on state of the substance?

Consider a collision in which the molecule with velocity $v_{x}$ strikes one face of the cube along x-axis. It rebounds elastically in the opposite direction having velocity $-v$ , its momentum changes from $mv$ to $-mv$ . Change in momentum along x-axis;

\Delta p = -mv_{x} - (mv_{x}) = -mv_{x} - mv_{x} = -2mv_{x}

The molecule travels a distance $2L$ between the consecutive collisions with the same wall of the cube. Hence, time between two consecutive collisions with one side of the cube is:

\frac{2L}{v_{x}} \quad \ldots (13.3)

(Fig. 13.2: Molecule moving with velocity $v$ )

The average force exerted by the wall of cube on the molecule can be found using Newton's second law of motion. We know that force is equal to the rate of change of momentum. The area of the wall is $L^{2}$ . Thus,

F = \frac{\text{Change in momentum}}{\text{Time}} = \frac{|\Delta p|}{\Delta t} = \frac{+2mv_{x}}{\dfrac{2L}{v_{x}}} = \frac{mv_{x}^{2}}{L} \quad \ldots (13.4)

This expression represents the force exerted on the container wall by one molecule in the x-direction. The positive sign in the change of momentum is due to the force of the molecule by the wall which is in opposite direction to the change in momentum as a result of the Newton's third law of motion.

\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{F}{L^{2}}

Using equation,

P = \frac{\dfrac{mv_{x}^{2}}{L}}{L^{2}}

P = \frac{mv_{x}^{2}}{L^{3}} = \frac{mv_{x}^{2}}{V} \quad \ldots (13.5)

This is the pressure exerted on container wall by one molecule.

The total force, we must add up the contributions of all the molecules that strike the wall, allowing for the possibility that they all have different speeds. Dividing the magnitude of the total force $F_{x}$ along x-axis by the area of the wall ( $L^{2}$ ), then gives the pressure $P$ on that wall:

P = \frac{m}{V}(v_{1x}^{2} + v_{2x}^{2} + v_{3x}^{2} + \dots + v_{Nx}^{2}) \quad \ldots (13.6)

where $v_{1x}^{2}$ is the velocity of particle 1 in the x-direction, and $v_{2x}^{2}$ the velocity of particle 2 in the x-direction and so on. If there are $N$ number of particles in the container, then total mass is the number of particles times mass of the particle (i.e. $M = mN$ ).

Brain teaser

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How does the Kinetic theory of gases relate to the Boyle's law and Charles's law?

As we know that velocity striking the wall is $v_{x}$ and after striking the wall it is $-v_{x}$ which would result in average velocity zero. This means simple average does not work here. In order to avoid this zero, we need to do the square of the velocities before averaging.

\langle v_{x}^{2}\rangle = \frac{v_{1x}^{2} + v_{2x}^{2} + v_{3x}^{2} + \dots + v_{Nx}^{2}}{N} = \frac{\sum_{i=1}^{N}v_{ix}^{2}}{N}

If we take the square root of the average square of the velocities, it is known as the root mean square (rms) velocity and is written as:

(v_{\mathrm{rms}})_{x} = \sqrt{\langle v_{x}^{2}\rangle} \quad \ldots (13.8)

Since the molecules are moving randomly in all directions, only one-third of their total velocity will be directed along any one Cartesian axis. So,

\langle v^{2}\rangle = \langle v_{x}^{2}\rangle + \langle v_{y}^{2}\rangle + \langle v_{z}^{2}\rangle \quad \ldots (13.9)

Since gas molecules are moving randomly and have the same average speed in all three directions. This means that;

\langle v_{x}^{2}\rangle = \langle v_{y}^{2}\rangle + \langle v_{z}^{2}\rangle \Rightarrow 3\langle v_{x}^{2}\rangle \quad \ldots (13.10)

Total average velocity is:

\langle v_{x}^{2}\rangle = \frac{\langle v^{2}\rangle}{3} \quad \ldots (13.11)

The pressure is written as:

P = \frac{N}{3V}m\langle v^{2}\rangle \quad \ldots (13.12)

Then

PV = \frac{1}{3}Nm\langle v^{2}\rangle \quad \ldots (13.13)

This is the pressure of the gas molecules on the walls. The pressure in terms of density can be written as:

\rho = \frac{\text{Total mass}}{\text{Volume}} = \frac{M}{V}

Rearranging the pressure, Eq. (13.13) for $P$ , and substituting the density

PV = \frac{1}{3}\rho\langle v^{2}\rangle \quad \ldots (13.14)

This is the pressure of the gas which depends on the density and root mean square of velocity.

Since

\rho = \frac{mN}{V}

P = \frac{mN}{3V}\langle v^{2}\rangle \quad \text{or} \quad PV = \frac{1}{3}Nm\langle v^{2}\rangle

P = \frac{2N}{3V} \left\langle \frac{1}{2}mv^{2} \right\rangle

P = \frac{2}{3}N_{o} \left\langle \frac{1}{2}mv^{2} \right\rangle \quad \left(\text{Here } \frac{N}{V} = N_{o}\right)

P = \frac{2}{3}N_{o}\langle\mathrm{K.E.}_{T}\rangle

Example 13.1

An ideal gas has a density of $4.5\,\mathrm{kg\,m^{-3}}$ at a pressure of $9.3 \times 10^{5}\,\mathrm{Pa}$ and a temperature of $504\,\mathrm{K}$ . Determine the root-mean-square speed of the gas atoms.

Solution

$\rho = 4.5\,\mathrm{kg\,m^{-3}}$ , $P = 9.3 \times 10^{5}\,\mathrm{Pa}$ , $v_{\mathrm{rms}} = ?$

Pressure is defined as; $P = \dfrac{1}{3}\rho\langle v^{2}\rangle$

\langle v^{2}\rangle = \frac{3P}{\rho}

\langle v^{2}\rangle = \frac{3 \times (9.3 \times 10^{5}\,\mathrm{Pa})}{4.5\,\mathrm{kg\,m^{-3}}} = 6.2 \times 10^{5}\,\mathrm{m^{2}\,s^{-2}}

To find the rms value, take the square root of the mean-square-speed, thus

v_{\mathrm{rms}} = \sqrt{\langle v^{2}\rangle} = \sqrt{6.2 \times 10^{5}\,\mathrm{m^{2}\,s^{-2}}} = 787\,\mathrm{m\,s^{-1}} \approx 7.9 \times 10^{2}\,\mathrm{m\,s^{-1}}

Example 13.2

What is the rms speed of air molecules ( $\mathrm{O}_{2}$ and $\mathrm{N}_{2}$ ) at room temperature e.g. $25^{\circ}\mathrm{C}$ ?

Solution

Mass of oxygen $\mathrm{O}_{2} = 32\,\mathrm{u}$ and Mass of nitrogen $\mathrm{N}_{2} = 28\,\mathrm{u}$

$1\,\mathrm{u} = 1.66 \times 10^{-27}\,\mathrm{kg}$ ; $k_{B} = 1.38 \times 10^{-23}\,\mathrm{J\,K^{-1}}$ ; $T = 25^{\circ}\mathrm{C} = 298\,\mathrm{K}$

Mass of oxygen $\mathrm{O}_{2} = 32 \times 1.66 \times 10^{-27}\,\mathrm{kg} = 5.3 \times 10^{-26}\,\mathrm{kg}$

Mass of Nitrogen $\mathrm{N}_{2} = 28 \times 1.66 \times 10^{-27}\,\mathrm{kg} = 4.6 \times 10^{-26}\,\mathrm{kg}$

The rms speed of oxygen is:

v_{\mathrm{rms}} = \sqrt{\frac{3k_{B}T}{m}} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}\,\mathrm{J\,K^{-1}}) \times 298\,\mathrm{K}}{5.3 \times 10^{-26}\,\mathrm{kg}}} = 481\,\mathrm{m\,s^{-1}} \approx 4.81 \times 10^{2}\,\mathrm{m\,s^{-1}}

Similarly, the rms speed of nitrogen is $v_{\mathrm{rms}} = 515\,\mathrm{m\,s^{-1}} \approx 5.15 \times 10^{2}\,\mathrm{m\,s^{-1}}$

The speed of air molecules is much greater than the speed of sound which is about $340\,\mathrm{m\,s^{-1}}$ .

13.3 Pressure in a Gas and Derivation of PV=13Nm⟨v2⟩PV = \tfrac{1}{3}Nm\langle v^{2}\ranglePV=31​Nm⟨v2⟩

Example 13.1

Example 13.2

Discussion & Community

13.3 Pressure in a Gas and Derivation of $PV = \tfrac{1}{3}Nm\langle v^{2}\rangle$