Question

The atmosphere of the sun consists mostly of hydrogen atoms (not molecules) at a temperature of 6000 K. What are (a) the average translational kinetic energy per atom and (b) the rms speed of the atoms

Answer 1

Explanation:

(a)  Formula for average translational kinetic energy of a particle is as follows.

                   U = $\frac{3}{2}(\frac{RT}{N})$

where,   R = Reydberg's constant

              T = absolute temperature

              N = Avogadro's number

Therefore, we will calculate value of average translational kinetic energy as follows.

                 U = $\frac{3}{2}(\frac{RT}{N})$

                    = $\frac{3}{2}(\frac{8.314 J/mol K \times 6000 K}{6.023 \times 10^{23} mol^{-1}})$

                    = $1.24 \times 10^{-19}$ J

Therefore, value of average translational kinetic energy is $1.24 \times 10^{-19}$ J.

(b)   Formula for average kinetic energy is as follows.

             K.E = $\frac{1}{2}(\frac{M}{N})v^{2}_{rms}$

Here,   M = molar mass = 1 kg/K mol

And, the average kinetic energy is equal to the average translational kinetic energy.

Hence,   K.E = U

     $\frac{1}{2}(\frac{M}{N})v^{2}_{rms}$ = $\frac{3}{2}(\frac{RT}{N})$

            $v^{2}_{rms} = \frac{3RT}{M}$

            $v = \sqrt{\frac{3RT}{M}}$

therefore, we will calculate r.m.s speed of the given atom as follows.

              $v = \sqrt{\frac{3RT}{M}}$

              $v = \sqrt{\frac{3 \times 8.314 J/mol K \times 6000 K}{1 kg/K mol}}$

              = 386.84 m/s

Hence, value of r.m.s speed of the given atom is 386.84 m/s.