Question

Kinetic theory of gases proves:
(A). Only Boyle’s law
(B). Only Charle’s law
(C). Only Avogadro’s law
(D). All of these

Answer 1

The kinetic theory of gases explains the thermodynamic behavior of gases with the help of which many principle concepts of thermodynamics were established. The model describes the gas as a large number of identical submicroscopic particles, all of which are in constant, rapid, random motion.

Complete step by step answer:
Kinetic molecular theory states that the gas particles are in constant motion and exhibit perfectly elastic collisions. The basic assumptions of this theory are as follows:
A. The volume occupied by the individual particles of a gas negligible as compared to the volume of gas itself.
B. The particles of an ideal gas exert no attractive forces on each other and on their surroundings.
C. Gas particles are in a constant state of random motion and move in straight lines until they do collide with another body.
D. The collisions exhibited by the gas particles are completely elastic i.e. when two molecules collide, total kinetic energy is conserved.
E. The average kinetic energy of gas molecules is directly proportional to absolute temperature only. This implies that all molecular motions seizes if the temperature is reduced to absolute zero.
Boyle’s law
According to Boyle’s law at constant temperature pressure and volume of gas are related as $PV = \operatorname{Constant}$
According to KTG, pressure of gas is
$P = \dfrac{1}{3}r{V^2}$
Since, $PV = \dfrac{1}{3}mN{V^2}$
Multiplying and dividing by $2$ on $R.H.S$
$PV = \dfrac{2}{3}N\left( {\dfrac{1}{2}m{V^2}} \right)$
$PV = \dfrac{2}{3}NKE = \operatorname{Constant}$
Hence, Boyle law is proved.
Charle’s law:
We know that, from KTG
$\left( {Pressure} \right){\text{ P = }}\dfrac{{mn{\vartheta ^2}}}{2}$
$\vartheta \to$ R.M.S. speed
Also $N = \dfrac{n}{V}$
($N \to$Total no. of gas molecules)
($V \to$ Volume)
$PV = \dfrac{{mN{V^2}}}{3}$
Now, $E = \dfrac{{m{V^2}}}{2}$(kinetic energy of a molecule)
Thus $PV = \dfrac{2}{3}NE$
Now from kinetic interpretation of temperature.
$E = \dfrac{3}{2}KT$ , $K \to$ Boltzmann constant
Thus, $PV = NKT$
Since $N$and $K$ are constants then for faxed $P$
$\dfrac{V}{T} = \operatorname{constant}$
So, Charle’s law is proved.
Avogadro’s law:
According to KTG,
$P = \dfrac{1}{3}\rho {\vartheta ^2} = \dfrac{1}{3}\dfrac{M}{V}{\vartheta ^2} = \dfrac{1}{3}\dfrac{{nm}}{V}{\vartheta ^2}$
For gas having mass ${m_1}$
$P = \dfrac{1}{3}\dfrac{{{n_1}{m_1}}}{V}\vartheta _1^2 - - - - \left( 1 \right)$
For gas having mass ${m_2}$
$P = \dfrac{1}{3}\dfrac{{{n_2}{m_2}}}{V}\vartheta _2^2 - - - - - - - \left( 2 \right)$
From first and second equation
${m_1}{n_1}\vartheta _1^2 = {m_2}{n_2}\vartheta _2^2{\text{ - - - - - }}\left( 3 \right)$
Since the temperature of both the gases are same,So
$\dfrac{1}{2}{m_1}\vartheta _1^2 = \dfrac{1}{2}{m_2}\vartheta _2^2{\text{ - - - - - - - - - }}\left( 4 \right)$
From equation $(3)$ and $(4)$
We can conclude that
${n_1} = {n_2}$
Number of molecules of both the gases is the same.
Hence, Avogadro’s law is proved.
Therefore the correct answer is option C.

Note: The individual molecules possess the standard physical properties of mass, momentum and energy. The density of the gas is simply the sum of the mass of the molecule divided by the volume which is occupied by the gas.