Respuesta :
The average translational kinetic energy of one molecule is [tex]\fbox{\begin\\1.71\times{10^{-20}}\,{\text{J}}\end{minispace}}[/tex].
The average translational kinetic energy of [tex]1\,{\text{mole}}[/tex] molecules is [tex]\fbox{\begin\\10297\,{\text{J}}\end{minispace}}[/tex].
Further Explanation:
The translation kinetic energy is the average kinetic energy possessed by a molecule as it moves inside the gas sample.
The average translational kinetic energy of a molecule is given by:
[tex]\fbox{\begin\\{K_E}=\dfrac{3}{2}{k_B}T\\\end{minispace}}[/tex]
Here, [tex]{K_E}[/tex] is the average kinetic energy, [tex]{k_B}[/tex] is the Boltzmann constant and T is the temperature of the gas sample.
The temperature of the gas is [tex]827\text{ K}}[/tex] and the value of Boltzmann constant is [tex]1.38\times{10^{-23}}\,{{\text{J}}\mathord{\left/{\vphantom{{\text{J}}{\text{K}}}}\right.\kern-\nulldelimiterspace}{\text{K}}}[/tex].
Substitute the values in above expression.
[tex]\begin{gathered}{K_E}=\frac{3}{2}\left({1.38\times{{10}^{-23}}}\right)\left({827}\right)\\=\frac{{3.423\times{{10}^{-20}}}}{2}\,{\text{J}}\\{\text{=1}}{\text{.71}}\times{\text{1}}{{\text{0}}^{-20}}\,{\text{J}}\\\end{gathered}[/tex]
Thus, the translational kinetic energy possessed by one molecule is [tex]1.71\times10^{-20}\text{ J}[/tex].
Now, we know that the number of molecules of a substance present in [tex]1\,{\text{mole}}[/tex] is [tex]6.022\times{10^{23}}[/tex] molecules.
So, we can calculate the average kinetic energy of one mole of gas molecules as the sum of energy of all the molecules.
[tex]{K_{total}} = n \cdot {K_E}[/tex]
Here, [tex]n[/tex] is the number of molecules in [tex]1\text{ mole}[/tex] and [tex]{K_{total}}[/tex] is the average kinetic energy of all the molecules.
Substitute the values.
[tex]\begin{gathered}{K_E}=\frac{3}{2}\left({1.38\times{{10}^{-23}}}\right)\left({827}\right)\\=\frac{{3.423\times{{10}^{-20}}}}{2}{\mkern1mu}{\text{J}}\\{\text{=1}}{\text{.71}}\times{\text{1}}{{\text{0}}^{-20}}{\mkern1mu}{\text{J}}\\\end{gathered}[/tex]
Thus, the average kinetic energy of [tex]1 \text{ mole}[/tex] of the gas is [tex]10297\,{\text{J}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Kinetic theory of gases
Keywords:
Kinetic energy, translational, average, 1 mole, 827 K, gas molecule, gas, single molecule, Boltzmann constant, temperature.
The average translational kinetic energy for one mole of gas at 827 k is [tex]1.711 \times 10 ^{-23} Joules[/tex]
The formula for calculating the average translational kinetic energy which is also called average kinetic energy is expressed as:
[tex]KE =\frac{3}{2}kT[/tex]
k is the Boltzmann constant
T is the temperature of the gas
Given the following expression:
k = 1.38× 10⁻²³ J/K
T = 827k
Substitute the given parameters into the formula as shown:
[tex]KE=\frac{3}{2} \times 1.38 \times 10^{-23}\times 827\\KE = \frac{3,423.78\times 10^{-23}}{2}\\KE= 1,711.89\times 10^{-23}\\KE=1.711\times 10^{-20}J\\[/tex]
Hence the average translational kinetic energy for one mole of gas at 827 k is [tex]1.711 \times 10 ^{-23} Joules[/tex]
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