Question

The average kinetic energy of one molecule of an ideal gas $27{}^{\circ }C$ and $1atm$ pressure is:
a.) $900$ $cal{K}^{-1}mo{l}^{-1}$
b.) $6.21\times {10}^{-21}J{K}^{-1}molecul{e}^{-1}$
c.) $336.7$ $J{K}^{-1}molecul{e}^{-1}$
d.) $3741.3$ $J{K}^{-1}mo{l}^{-1}$

Answer 1

Hint: First find out the data which we have given set according to the units required then by the help of Kinetic theory of gas we will find the average kinetic energy per molecule. Here if a molecule has variable speed then it must have a variable kinetic energy too under these circumstances, we can conclude only about average kinetic energy.

Complete answer:
First of all, we know that by the help of kinetic theory of gases we can analyse the theoretical model for all the experimental gas laws which is based on some assumption some of which are:
The average kinetic energy of gas depends only on its absolute temperature, the molecule moves with different speed however the speed of each molecule keeps on changing due to collision, there is no interaction between gaseous particles well the interaction may appreciably affect under certain conditions like temperature and pressure.
So now returning to our question
To convert the temperature given in centigrade to kelvin we will do the following:
$T=27+273=300K$
And now let’s find the Average Kinetic energy per molecule which is
${\displaystyle \frac{3}{2}}{\displaystyle \frac{RT}{N_0}}={\displaystyle \frac{3}{2}}KT$
Here, $R=$Gas constant
$T=$Temperature
$N_0=$Avogadro Number
$K=$Boltzmann’s constant
Now by placing the values we will get
$\frac{3}{2}\frac{RT}{N_0}={\displaystyle \frac{8.314}{6.023\times {10}^{23}}}\times 300$
Here,
$\begin{array}{rl}& R=8.314J/molK\ast \\ & T=300K\\ & N_0=6.023\times {10}^{23}\end{array}$
          $=6.21\times {10}^{-21}J{K}^{-1}molecul{e}^{-1}$
So we get our answer as
$=6.21\times {10}^{-21}J{K}^{-1}molecul{e}^{-1}$.

Note: In kinetic theory it is assumed that average kinetic energy of the gas molecule is dependent on the absolute temperature.
The average kinetic energy is dependent on the absolute temperature which we use in the above solution.
If $T=0K$ (i.e., $-273.15{}^{\circ }C$),then average kinetic energy is zero so we can say that absolute zero is the temperature at which molecular motion ceases.