Question

The kinetic energy per molecule of a gas at temperature $T$ is ____________.

• A. $\left(\frac{3}{2}\right)RT$
• B. $\left(\frac{3}{2}\right)K_{B}T$
• C. $\left(\frac{2}{3}\right)RT$
• D. $\left(\frac{3}{2}\right)\left(\frac{RT}{M}\right)$

Answer 1

The correct option is B $\left(\frac{3}{2}\right)K_{B}T$

We know that for any gas kept in a chamber, its pressure is given by

$P=\frac{1}{3}\varrho \times C_{RMS}^2$

Here $\varrho$ is the density i.e. $\frac{M}{V}$

$P=\frac{M{C}_{{}_{RMS}}^2}{3V}$

$\therefore 3PV=M{C}_{{}_{RMS}}^2$

For ideal gas, $PV=nRT$ n is no of moles, R is universal gas constant, T is temperature.

$\therefore \frac{3}{2}nRT=\frac{1}{2}M{C}_{{}_{RMS}}^2------$Kinetic Energy

Remember that 1 mole means 1 Avogadro number i.e.$N_A$

$\therefore$ KE per molecule $=\frac{3}{2}\frac{RT}{N_A}$

$\frac{R}{N_A}=K_B$--------Boltzmann constant

$\therefore K.E.$ $per$ $molecule$ $=\frac{3}{2}{K}_{B}T$