Question

Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is

• A. $\left[\frac{1}{3}\right]^{rd}$ kinetic energy per unit volume of a gas
• B. $\left[\frac{2}{3}\right]^{rd}$ kinetic energy per unit volume of a gas
• C. $\left[\frac{3}{4}\right]^{th}$ kinetic energy per unit volume of a gas
• D. $\frac{3}{3} \times$ kinetic energy per unit volume of a gas

Answer 1

Answer: (B)

$\left[\frac{2}{3}\right]^{rd}$ kinetic energy per unit volume of a gas

Answer 2

The kinetic energy of the gas is given by

$K =\frac{1}{2} mv _{ rms }^{2}$

Where

• $v_{rms}$ is the root mean square (rms) velocity of the molecules of the gas
• $m$ is the mass of one molecule of the gas

$R m s$ velocity of gas molecules is given by

$v _{ rms }=\sqrt{\frac{3 R T}{m}}$

Where

• T is the absolute temperature
• R is the Universal gas constant

$\therefore K =\frac{1}{2} m \cdot \frac{3 RT }{ m }=\frac{3}{2} RT$

From the ideal gas equation, $RT = PV$
$\Rightarrow K =\frac{3}{2} PV$

Thus we get the pressure of the gas $P =\frac{2}{3} \frac{ K }{ V }$

Therefore, the pressure exerted by the gas on the walls of the container is $[\frac{2}{3}]^{rd}$ kinetic energy per unit volume of a gas.

_Discover More from Chapter: _Kinetic Theory of Gases