Respuesta :
Explanation:
(a) Formula for average translational kinetic energy of a particle is as follows.
U = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
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 = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
= [tex]\frac{3}{2}(\frac{8.314 J/mol K \times 6000 K}{6.023 \times 10^{23} mol^{-1}})[/tex]
= [tex]1.24 \times 10^{-19}[/tex] J
Therefore, value of average translational kinetic energy is [tex]1.24 \times 10^{-19}[/tex] J.
(b) Formula for average kinetic energy is as follows.
K.E = [tex]\frac{1}{2}(\frac{M}{N})v^{2}_{rms}[/tex]
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
[tex]\frac{1}{2}(\frac{M}{N})v^{2}_{rms}[/tex] = [tex]\frac{3}{2}(\frac{RT}{N})[/tex]
[tex]v^{2}_{rms} = \frac{3RT}{M}[/tex]
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
therefore, we will calculate r.m.s speed of the given atom as follows.
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
[tex]v = \sqrt{\frac{3 \times 8.314 J/mol K \times 6000 K}{1 kg/K mol}}[/tex]
= 386.84 m/s
Hence, value of r.m.s speed of the given atom is 386.84 m/s.
Following are the solution to the given points:
Given:
Temperature [tex]= 6000\ K[/tex]
To find:
kinetic energy =?
rms speed of the atoms=?
Solution:
For point a)
[tex]\to Avg \ KE =\frac{3}{2}\ KT \\\\[/tex]
[tex]\to KE_{avg} =1.5 \times 1.38 \times 10^{-23} \times 6000 \\\\[/tex]
[tex]=1.5 \times 1.38 \times 10^{-23} \times 6000 \\\\=2.07 \times 10^{-23} \times 6000 \\\\=2.07 \times 10^{-23} \times 6000 \\\\=12.420 \times 10^{-20}\ J[/tex]
For point b)
[tex]\to[/tex] rms speed [tex]V_{rms}[/tex]:
[tex]\to v_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
[tex]=\sqrt{\frac{3 \times 8314 \times 6000}{1}} \\\\=\sqrt{\frac{24942\times 6000}{1}} \\\\=\sqrt{149652000}\\\\=12233.233\\\\ = 12233\ \frac{m}{s}[/tex]
Therefore, the final answer is "[tex]12.420 \times 10^{-20} \ J\ \ and \ \ 12233 \ \frac{m}{s}[/tex]".
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