Chapter 13, Problem 59P

To determine

To show:The pressure of the gas in terms of the density and root mean square speed of the molecule.

  $P=\frac{1}{3}\rho v^2{}_{rms}$

Explanation of Solution

Formula used:

  $P=\frac{NkT}{V}$

Here,

  $P$ = Pressure

  $V$ = Volume

  $N$ = No. of moles

  $k$ = Boltzmann constant

  $T$ = Temperature

Mass of the gas = number of the molecule*Mass of the molecule

  $M=Nm$

Density of gas

  $\rho =\frac{M}{V}$

Calculation:

  $\begin{array}{l}P=\frac{NkT}{V}\mathrm{.....................}(1)\\ v_{rms}=\sqrt{\frac{3kT}{m}}\to T=m{v}^2{}_{rms}/3k\mathrm{................}(2)\end{array}$

Mass of the gas = number of the molecule*Mass of the molecule

  $M=Nm\mathrm{............}(3)$

Density of gas

  $\rho =\frac{M}{V}\mathrm{...................}(4)$

Use equation (2), (3) and (4) in (1)

  $\begin{array}{l}P=\frac{NkT}{V}=\frac{Nk}{V}\frac{m{v}^2{}_{rms}}{3k}\\ =\frac{1}{3}\frac{Nm}{V}{v}^2{}_{rms}=\frac{1}{3}\frac{M}{V}{v}^2{}_{rms}\\ =\frac{1}{3}\rho v^2{}_{rms}\\ P=\frac{1}{3}\rho v^2{}_{rms}\end{array}$

Conclusion:

The pressure of the gas is the $P=\frac{1}{3}\rho v^2{}_{rms}$