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)

Kinetic energy of the gas $K=\frac{1}{2}m{v}_{rms}^2$
$Rms$ velocity of gas molecules $v_{rms}=\sqrt{\frac{3RT}{m}}$ $\therefore K=\frac{1}{2}m\cdot \frac{3RT}{m}=\frac{3}{2}RT$
From ideal gas equation, $RT=PV$
$\Rightarrow K=\frac{3}{2}PV$
Thus we get pressure of the gas $P=\frac{2}{3}\frac{K}{V}$