Calculate the average translational kinetic energy (sometimes just called average kinetic energy)â¦ for one mole of gas at 827 k

Question

MrMuchimi

4 years ago

4

Calculate the average translational kinetic energy (sometimes just called average kinetic energy)â¦ for one mole of gas at 827 k

. and... for a single gas molecule at 827 k.

Physics

2 answers:

andrew11 [14] · Answer 1 · 2021-11-26T09:53:04+03:00

1) Let's calculate the average kinetic energy of a single molecule first. This is given by:
$E= \frac{3}{2} k_B T$
where $k_B = 1.38 \cdot 10^{-23} J/K$ is the Boltzmann's constant, while T is the temperature in Kelvin. Using T=827 K, we find
$E= \frac{3}{2}(1.38 \cdot 10^{-23} J/K) (827 K)=1.7 \cdot 10^{-20}J$

2) To find the average kinetic energy for one mole of gas, let's keep in mind that 1 mole of gas contains a number of molecules equal to Avogadro's number:
$N(1 mol)=6.022 \cdot 10^{23} mol^{-1}$
So, to find the average kinetic energy of the whole gas, we must multiply the energy we found at point 1) by the number of molecules of the gas, which is $N(1 mol)\cdot 1 mol$ :
$E_{gas}=6.022 \cdot 10^{23} mol^{-1}\cdot 1 mol\cdot 1.7 \cdot 10^{-20}J=10237 J$

dimulka [17.4K] · Answer 2 · 2021-11-26T11:02:47+03:00

The average translational kinetic energy of one molecule is $\fbox{\begin\\1.71\times{10^{-20}}\,{\text{J}}\end{minispace}}$ .

The average translational kinetic energy of $1\,{\text{mole}}$ molecules is $\fbox{\begin\\10297\,{\text{J}}\end{minispace}}$ .

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:

$\fbox{\begin\\{K_E}=\dfrac{3}{2}{k_B}T\\\end{minispace}}$

Here, ${K_E}$ is the average kinetic energy, ${k_B}$ is the Boltzmann constant and T is the temperature of the gas sample.

The temperature of the gas is $827\text{ K}}$ and the value of Boltzmann constant is $1.38\times{10^{-23}}\,{{\text{J}}\mathord{\left/{\vphantom{{\text{J}}{\text{K}}}}\right.\kern-\nulldelimiterspace}{\text{K}}}$ .

Substitute the values in above expression.

$\begin{gathered}{K_E}=\frac{3}{2}\left({1.38\times{{10}^{-23}}}\right)\left({827}\right)\\=\frac{{3.423\times{{10}^{-20}}}}{2}\,{\text{J}}\\{\text{=1}}{\text{.71}}\times{\text{1}}{{\text{0}}^{-20}}\,{\text{J}}\\\end{gathered}$

Thus, the translational kinetic energy possessed by one molecule is $1.71\times10^{-20}\text{ J}$ .

Now, we know that the number of molecules of a substance present in $1\,{\text{mole}}$ is $6.022\times{10^{23}}$ molecules.

So, we can calculate the average kinetic energy of one mole of gas molecules as the sum of energy of all the molecules.

${K_{total}} = n \cdot {K_E}$

Here, $n$ is the number of molecules in $1\text{ mole}$ and ${K_{total}}$ is the average kinetic energy of all the molecules.

Substitute the values.

$\begin{gathered}{K_E}=\frac{3}{2}\left({1.38\times{{10}^{-23}}}\right)\left({827}\right)\\=\frac{{3.423\times{{10}^{-20}}}}{2}{\mkern1mu}{\text{J}}\\{\text{=1}}{\text{.71}}\times{\text{1}}{{\text{0}}^{-20}}{\mkern1mu}{\text{J}}\\\end{gathered}$