Question

Question: (II)Show that the rms speed of molecules in a gas is given by\({{\bf{v}}_{{\bf{rms}}}}{\bf{ = }}\sqrt {{\bf{3P/\rho }}} \)where P is the pressure in the gas and\({\bf{\rho }}\)is the gas density.

Short answer

It is proved that the rms speed of molecule in a gas is \({v_{{\rm{rms}}}} = \sqrt {3P/\rho } \).

Step-by-step solution

Understanding the ideal gas law

Ideal gas law states that at constant volume, the pressure is directly proportional to the number of molecules and temperature of the gas.

The ideal gas equation is written as follows:

\(PV = nkT\)

Here, P is the pressure, V is the volume, n is the number of molecules, k is the Boltzmann constant, and T is the temperature.

Determination of the temperature and density of the gas

For one molecule, the ideal gas equation becomes

\(\begin{aligned}{c}PV &= \left( 1 \right)kT\\PV &= kT\end{aligned}\)

The density of a gas is given by the following:

\(\rho = \frac{m}{V}\)

Here, mis the mass of the gas.

Evaluation of the rms speed of the molecule in a gas

The relation of rms speed can be written as follows:

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3kT}}{m}} \)

Substitute the values in the above equation.

\(\begin{aligned}{l}{v_{{\rm{rms}}}} &= \sqrt {\frac{{3PV}}{{\rho V}}} \\{v_{{\rm{rms}}}} &= \sqrt {\frac{{3P}}{\rho }} \end{aligned}\)

Thus, it is proved that the rms speed of molecule in a gas is \({v_{{\rm{rms}}}} = \sqrt {3P/\rho } \).