Chapter 13, Problem 77GP

To determine

To Find: The RMS speed of the hydrogen and the pressure of the hydrogen.

Answer to Problem 77GP

The rms speed of the hydrogen atoms is $260\text{ m/s}$ and the pressure of the hydrogen atom is $3.7\times {10}^{-22}\text{ atm}$ .

Explanation of Solution

Given:

Density of water ${\rho }_{matter}=1{\text{ atom / cm}}^{\text{3}}$ .

Temperature $T=2.7\text{ K}$ .

Mass of $1$ hydrogen atom $m=1\times 1.66\times {10}^{-27}\text{ kg}$ .

Boltzmann constant $k=1.38\times {10}^{-23}\text{ J/K}$ .

Formula used:

The RMS speed of the hydrogen atom is obtained by the following formula,

  $v_{rms}=\sqrt{\frac{3kT}{m}}$

And for the pressure of the hydrogen atom

The pressure of the hydrogen atom is obtained by the following formula,

  $\text{Pressure }P=\frac{N\times k\times T}{V}$

Calculation:

Substitute the given values in the above equation,

  $v_{rms}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 2.7}{1\times 1.66\times {10}^{-27}}}=260\text{ m/s}$

For the pressure of the hydrogen atom

Substitute the given values in the above equation,

  $\begin{array}{l}\text{Pressure }P=\frac{1\times 1.38\times {10}^{-23}\times 2.7}{{10}^{-6}}=3.726\times {10}^{17}\text{ Pa}\\ P=3.726\times {10}^{17}\text{ Pa }\times \frac{1\text{ atm}}{1.01\times {10}^5\text{ Pa}}=3.7\times {10}^{-22}\text{ atm}\end{array}$

Conclusion:

Thus, the RMS speed of the hydrogen atoms is $260\text{ m/s}$ . pressure of the hydrogen atom is $3.7\times {10}^{-22}\text{ atm}$ .