Question
State the assumptions of the kinetic theory of gases. The density of carbon dioxide gas at $0^{\circ} \mathrm{C}$ and at a pressure of $1.0 \times 10^5$ newton/ metre ${ }^2$ is $1.98 \mathrm{~kg} / \mathrm{m}^3$. Find the root mean square velocity of its molecules at $0^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. Pressure is constant.

Answer

The entire structure of the kinetic theory of gases is based on the following assumptions which were first stated by Classius.
  1. A gas consists of a very large number of molecules (of the order of Avogadro's number, $10^{23}$), which are perfect elastic spheres. They are identical in all respects for a given gas and are different for different gases.
  2. The molecules of a gas are in a state of incessant random motion. They move in all directions with different speeds, (of the order of 500m/s) and obey Newton's laws of motion.
  3. The size of the gas molecules is very small as compared to the distance between them. If typical size of a molecule is $2\mathring{\text{A}},$ average distance between the molecules is $\geq20\mathring{\text{A}}.$ Hence volume occupied by the molecules is negligible in comparison to the volume of the gas.
  4. The molecules do not exert any force of attraction or repulsion on each other, except during collision.
  5. The collisions of the molecules with themselves and with the walls of the vessel are perfectly elastic. As such the momentum and the kinetic energy of the molecules are conserved during collisions, though their velocities change.
  6. There is no concentration of the molecules at any point inside the container i.e., molecular density is uniform throughout the gas.
  7. A molecule moves along a straight line between two successive collisions and the average straight distance covered between two successive collisions is called the mean free path of the molecules.
  8. The collisions are almost instantaneous, i.e., the time of collision of two molecules is negligible as compared to time interval between two successive collisions.
Numerical: We know that$\text{P}=\frac{1}{3}\rho\text{v}^2$
$\therefore\text{v}_{\text{rms}}=\sqrt{\big(\text{v}^2\big)}=\sqrt{\Big(\frac{3\text{P}}{\rho}\Big)}$
Given that $\text{P}=1.0\times10^5\text{newton/metre}^2$$\rho=1.98\text{kg/metre}^3$
$\therefore\text{v}_{\text{rms}}=\sqrt{\Big[\Big\{\frac{3\times(1.0\times10^5)}{1.98}\Big\}\Big]}$
$=389\text{metre/sec}.$
From kinetic theory of gases, the root mean square speed is directly proportional to the square root of absolute temperature$\text{v}_{\text{rms}}\propto\sqrt{\text{T}}$
$\therefore\frac{(\text{v}_{\text{rms}})_{30}}{(\text{v}_{\text{rms}})_0}=\sqrt{\Big[\Big(\frac{273+30}{273+0}\Big)\Big]}$
$=\sqrt{\Big[\Big(\frac{303}{273}\Big)\Big]}=1.053$
$(\text{v}_{\text{rms}})_{30}=(\text{v}_{\text{rms}})_0\times1.053$
$=389\times10.53=410\text{ metre/sec}$

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