Question
Give the postulates of kinetic theory of gases. Derive the expression for pressure exerted by gas molecules in a container. Use it to relate kinetic energy with pressure.
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Answer
The kinetic theory of gases is based on the following assumptions:
Consider an ideal gas contained in a cubical container OPQRSTKL of each side a and having a volume V.
Clearly, volume of the gas, V = volume of the container = $a^3$, i.e. $V = a^3$.
Let there be n molecules of the gas in the container each of mass m. Then, the total mass of the gas in the container is given by
$M = m \times n$
Let the random velocities of the gas molecules $A_1, A_2, .... A_n,$ be $c_1, c_2, ..... c_n$ respectively. Let $(x_1, y_1, z_1), (x_2, y_2, z_2), ..... (x_n, y_n, z_n)$ be the rectangular components of the velocities $c_1, c_2, ... c_n$, respectively, along three mutually perpendicular directions Ox, OY and Oz.
$\begin{matrix}\text{x}^2_1+\text{y}^2_1+\text{z}^2_1=\text{c}^2_1\\\text{x}^2_2+\text{y}^2_2+\text{z}^2_2=\text{c}^2_2\\.................\\\text{x}^2_\text{n}+\text{y}^2_\text{n}+\text{z}^2_\text{n}=\text{c}^2_\text{n}\end{matrix}$
The change in momentum of $A_1$, along OX after one collision = $-mx_1 - mx_1 --2mx_1$
According to law of conservation of linear momentum, in one collision, the momentum transferred to the wall QRSL by the molecule $A_1$ will be = $+2mx_1$
The molecule $A_1$ returned from face QRSL, collides against the wall OPKT, rebounds and again collides with QRSL. Thus, the molecule $A_1$ covers a distance 2a between two successive collisions with the wall QRSL.
$\therefore$ Time between two successive collisions, $\text{t}=\frac{\text{distance}}{\text{speed}}=\frac{2\text{a}}{\text{x}_1}$
It means the molecule $A_1$ will collide with wall QRSL after every $\frac{2\text{a}}{\text{x}_1}$ seconds.
Therefore, number of collisions per second with wall QRSL $=\frac{1}{\text{t}}=\frac{\text{x}_1}{2\text{a}}$
$\therefore$ Momentum transferred to the wall QRSL in one second by molecule $A_1$ = momentum transferred in one collision × no. of collisions in 1s.
$=2\text{mx}_1\times\frac{\text{x}_1}{2\text{a}}=\frac{\text{mx}^2_1}{\text{a}}$
According to Newton's 2nd law of motion, the rate of change of momentum of a body is equal to the force exerted on it. Hence, the force exerted by the molecule A, on the wall QRSL will be.
$\text{f}_1=\frac{\text{mx}^2_1}{\text{a}}$
$\text{f}_2=\frac{\text{mx}^2_2}{\text{a}},...\text{f}_\text{n}=\frac{\text{mx}^2_{\text{n}}}{\text{a}}$
Total force exerted by all the molecules on wall QRSL will be
$\text{F}_{\text{x}}=\frac{\text{mx}^2_1}{\text{a}}+\frac{\text{mx}^2_2}{\text{a}}+...+\frac{\text{mx}^2_\text{n}}{\text{a}}=\frac{\text{m}}{\text{a}}\big(\text{x}^2_1+\text{x}^2_2+\text{x}^2_{\text{n}}\big)$
$\therefore$ Pressure exerted on the wall QRSL is
$\text{P}_{\text{x}}=\frac{\text{force}}{\text{area of wall QRSL}}=\frac{\text{F}_{\text{x}}}{\text{a}^2}=\frac{\text{m}}{\text{a}^3}\big(\text{x}^2_1+\text{x}^2_2+...+\text{x}^2_\text{n}\big)$
Similarly, the pressure exerted by the gas molecules perpendicular to OY and OZ respectively are given by
$\text{P}_{\text{y}}=\frac{\text{m}}{\text{a}^3}\big(\text{y}^2_1+\text{y}^2_2+...+\text{y}^2_{\text{n}}\big)$
and $\text{P}_{\text{z}}=\frac{\text{m}}{\text{a}^3}\big(\text{z}^2_1+\text{z}^2_2+...+\text{z}^2_{\text{n}}\big)$
Since the molecular density is uniform throughout the gas, therefore the pressure exerted by the molecules is the same in all directions. Hence,
$\text{P}_{\text{x}}=\text{P}_{\text{y}}=\text{P}_{\text{z}}=\text{p}$ (say)
$\therefore\text{P}_{\text{x}}+\text{P}_{\text{y}}+\text{P}_{\text{z}}=3\text{P}$
or $\text{P}=\frac{\text{P}_{\text{x}}+\text{P}_{\text{y}}+\text{P}_{\text{z}}}{3}$
$\text{P}=\frac{1}{3}\Big[\frac{\text{m}}{\text{a}^3}\big(\text{x}^2_1+\text{x}^2_2+...+\text{x}^2_\text{n}\big)+\frac{\text{m}}{\text{a}^3}\big(\text{y}^2_1+\text{y}^2_2+...+\text{y}^2_{\text{n}}\big)\\+\frac{\text{m}}{\text{a}^3}\big(\text{z}^2_1+\text{z}^2_2+...+\text{z}^2_{\text{n}}\big)\Big]$
$=\frac{\text{m}}{3\text{a}^3}\Big[\big(\text{x}^2_1+\text{x}^2_2+...+\text{x}^2_{\text{n}}\big)+\big(\text{y}^2_1+\text{y}^2_2+...+\text{y}^2_{\text{n}}\big)\\+\big(\text{z}^2_1+\text{z}^2_2+...+\text{z}^2_{\text{n}}\big)\Big]$
$\text{P}=\frac{\text{m}}{3\text{a}^3}\Big[\big(\text{x}^2_1+\text{y}^2_1+\text{z}^2_1\big)+\big(\text{x}^2_2+\text{y}^2_2+\text{z}^2_2\big)\\+...+\big(\text{x}^2_\text{n}+\text{y}^2_\text{n}+\text{z}^2_\text{n}\big)\Big]$
$=\frac{\text{m}}{3\text{V}}\big[\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_\text{n}\big]=\frac{1}{3}\frac{\text{mv}}{\text{V}}\Big[\frac{\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_\text{n}}{\text{n}}\Big]$
$\text{P}=\frac{\text{M}}{3\text{V}}\text{c}^2$
where $\text{c}=\sqrt{\frac{\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_\text{n}}{\text{n}}}$ is called the root mean square (r.m.s.) velocity of the gas molecules.
Relation of K.E. with pressure:
Pressure $\text{P}=\frac{1}{3}\frac{\text{nmc}^2}{\text{V}}=\frac{1}{3}\frac{\text{M}}{\text{V}}\text{c}^2$
$\Rightarrow\text{P}=\frac{1}{3}\rho\text{c}^2$ $\Big[\because\frac{\text{M}}{\text{V}}=\rho\Big]$
Mean K.E. of translation per unit volume of gas is E $=\frac{1}2{}\rho\text{c}^2$
$\therefore\frac{\text{P}}{\text{E}}=\frac{\frac{1}{3}\rho\text{c}^2}{\frac{1}{2}\rho\text{c}^2}=\frac{2}{3}$
$\text{P}=\frac{2}{3}\text{E}$
- A gas consists of a very large number of molecules which are perfect elastic spheres and identical in all respects for a given gas. These are different for different gases.
- The molecules of a gas are in a state of continuous, rapid and random motion in all directions with different speeds, ranging from zero to infinity and obey Newton's laws of motion.
- The size of the gas molecules is very small as compared to the distance between them. Hence, the volume occupied by the molecules, is negligible in comparison to the volume of the gas.
- The molecules do not exert any force of attraction or repulsion on each other, except during collision.
- The collisions of the molecules with themselves and with the walls of the vessels are perfectly elastic, i.e. the momentum and the kinetic energy of the molecules are conserved during collisions.
Consider an ideal gas contained in a cubical container OPQRSTKL of each side a and having a volume V.
Clearly, volume of the gas, V = volume of the container = $a^3$, i.e. $V = a^3$.
Let there be n molecules of the gas in the container each of mass m. Then, the total mass of the gas in the container is given by
$M = m \times n$
Let the random velocities of the gas molecules $A_1, A_2, .... A_n,$ be $c_1, c_2, ..... c_n$ respectively. Let $(x_1, y_1, z_1), (x_2, y_2, z_2), ..... (x_n, y_n, z_n)$ be the rectangular components of the velocities $c_1, c_2, ... c_n$, respectively, along three mutually perpendicular directions Ox, OY and Oz.
$\begin{matrix}\text{x}^2_1+\text{y}^2_1+\text{z}^2_1=\text{c}^2_1\\\text{x}^2_2+\text{y}^2_2+\text{z}^2_2=\text{c}^2_2\\.................\\\text{x}^2_\text{n}+\text{y}^2_\text{n}+\text{z}^2_\text{n}=\text{c}^2_\text{n}\end{matrix}$
The change in momentum of $A_1$, along OX after one collision = $-mx_1 - mx_1 --2mx_1$
According to law of conservation of linear momentum, in one collision, the momentum transferred to the wall QRSL by the molecule $A_1$ will be = $+2mx_1$
The molecule $A_1$ returned from face QRSL, collides against the wall OPKT, rebounds and again collides with QRSL. Thus, the molecule $A_1$ covers a distance 2a between two successive collisions with the wall QRSL.
$\therefore$ Time between two successive collisions, $\text{t}=\frac{\text{distance}}{\text{speed}}=\frac{2\text{a}}{\text{x}_1}$
It means the molecule $A_1$ will collide with wall QRSL after every $\frac{2\text{a}}{\text{x}_1}$ seconds.
Therefore, number of collisions per second with wall QRSL $=\frac{1}{\text{t}}=\frac{\text{x}_1}{2\text{a}}$
$\therefore$ Momentum transferred to the wall QRSL in one second by molecule $A_1$ = momentum transferred in one collision × no. of collisions in 1s.
$=2\text{mx}_1\times\frac{\text{x}_1}{2\text{a}}=\frac{\text{mx}^2_1}{\text{a}}$
According to Newton's 2nd law of motion, the rate of change of momentum of a body is equal to the force exerted on it. Hence, the force exerted by the molecule A, on the wall QRSL will be.
$\text{f}_1=\frac{\text{mx}^2_1}{\text{a}}$
$\text{f}_2=\frac{\text{mx}^2_2}{\text{a}},...\text{f}_\text{n}=\frac{\text{mx}^2_{\text{n}}}{\text{a}}$
Total force exerted by all the molecules on wall QRSL will be
$\text{F}_{\text{x}}=\frac{\text{mx}^2_1}{\text{a}}+\frac{\text{mx}^2_2}{\text{a}}+...+\frac{\text{mx}^2_\text{n}}{\text{a}}=\frac{\text{m}}{\text{a}}\big(\text{x}^2_1+\text{x}^2_2+\text{x}^2_{\text{n}}\big)$
$\therefore$ Pressure exerted on the wall QRSL is
$\text{P}_{\text{x}}=\frac{\text{force}}{\text{area of wall QRSL}}=\frac{\text{F}_{\text{x}}}{\text{a}^2}=\frac{\text{m}}{\text{a}^3}\big(\text{x}^2_1+\text{x}^2_2+...+\text{x}^2_\text{n}\big)$
Similarly, the pressure exerted by the gas molecules perpendicular to OY and OZ respectively are given by
$\text{P}_{\text{y}}=\frac{\text{m}}{\text{a}^3}\big(\text{y}^2_1+\text{y}^2_2+...+\text{y}^2_{\text{n}}\big)$
and $\text{P}_{\text{z}}=\frac{\text{m}}{\text{a}^3}\big(\text{z}^2_1+\text{z}^2_2+...+\text{z}^2_{\text{n}}\big)$
Since the molecular density is uniform throughout the gas, therefore the pressure exerted by the molecules is the same in all directions. Hence,
$\text{P}_{\text{x}}=\text{P}_{\text{y}}=\text{P}_{\text{z}}=\text{p}$ (say)
$\therefore\text{P}_{\text{x}}+\text{P}_{\text{y}}+\text{P}_{\text{z}}=3\text{P}$
or $\text{P}=\frac{\text{P}_{\text{x}}+\text{P}_{\text{y}}+\text{P}_{\text{z}}}{3}$
$\text{P}=\frac{1}{3}\Big[\frac{\text{m}}{\text{a}^3}\big(\text{x}^2_1+\text{x}^2_2+...+\text{x}^2_\text{n}\big)+\frac{\text{m}}{\text{a}^3}\big(\text{y}^2_1+\text{y}^2_2+...+\text{y}^2_{\text{n}}\big)\\+\frac{\text{m}}{\text{a}^3}\big(\text{z}^2_1+\text{z}^2_2+...+\text{z}^2_{\text{n}}\big)\Big]$
$=\frac{\text{m}}{3\text{a}^3}\Big[\big(\text{x}^2_1+\text{x}^2_2+...+\text{x}^2_{\text{n}}\big)+\big(\text{y}^2_1+\text{y}^2_2+...+\text{y}^2_{\text{n}}\big)\\+\big(\text{z}^2_1+\text{z}^2_2+...+\text{z}^2_{\text{n}}\big)\Big]$
$\text{P}=\frac{\text{m}}{3\text{a}^3}\Big[\big(\text{x}^2_1+\text{y}^2_1+\text{z}^2_1\big)+\big(\text{x}^2_2+\text{y}^2_2+\text{z}^2_2\big)\\+...+\big(\text{x}^2_\text{n}+\text{y}^2_\text{n}+\text{z}^2_\text{n}\big)\Big]$
$=\frac{\text{m}}{3\text{V}}\big[\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_\text{n}\big]=\frac{1}{3}\frac{\text{mv}}{\text{V}}\Big[\frac{\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_\text{n}}{\text{n}}\Big]$
$\text{P}=\frac{\text{M}}{3\text{V}}\text{c}^2$
where $\text{c}=\sqrt{\frac{\text{c}^2_1+\text{c}^2_2+...+\text{c}^2_\text{n}}{\text{n}}}$ is called the root mean square (r.m.s.) velocity of the gas molecules.
Relation of K.E. with pressure:
Pressure $\text{P}=\frac{1}{3}\frac{\text{nmc}^2}{\text{V}}=\frac{1}{3}\frac{\text{M}}{\text{V}}\text{c}^2$
$\Rightarrow\text{P}=\frac{1}{3}\rho\text{c}^2$ $\Big[\because\frac{\text{M}}{\text{V}}=\rho\Big]$
Mean K.E. of translation per unit volume of gas is E $=\frac{1}2{}\rho\text{c}^2$
$\therefore\frac{\text{P}}{\text{E}}=\frac{\frac{1}{3}\rho\text{c}^2}{\frac{1}{2}\rho\text{c}^2}=\frac{2}{3}$
$\text{P}=\frac{2}{3}\text{E}$
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