Question

Calculate –
The average kinetic energy of translation of an oxygen molecule at \[{27^0}C\].
The total kinetic energy of an oxygen molecule at \[{27^0}C\] .
The total kinetic energy in joule of one mole of oxygen at \[{27^0}C\] .
Given Avogadro’s number $ = 6.02 \times {10^{23}}$ and Boltzmann’s constant $ = 1.38 \times {10^{ - 23}}J/(mol - K)$ .

Answer 1

Hint: We have studied in the chapter Kinetic Theory of gases about kinetic energy of molecules so we know that the average kinetic energy and total kinetic energy is a function of temperature. In average kinetic energy we used translational degrees of freedom while in the calculation of total kinetic energy we use total degrees of freedom.

Complete step-by-step solution -
There are three translational degrees of freedom, thus the average kinetic energy of an oxygen molecule at \[{27^0}C\] is given by the formula ,${E_T} = \dfrac{3}{2}kT$ where ${E_T}$ is average kinetic energy, $k$ is Boltzmann’s constant which is equals to $1.38 \times {10^{ - 23}}J/(mol - K)$ and $T$ is temperature.
So lets calculate the average kinetic energy of translation of an oxygen molecule at \[{27^0}C\] ;
  $ \Rightarrow {E_T} = \dfrac{3}{2} \times 1.38 \times {10^{ - 23}} \times 300$ (here $T = (27 + 273)K$)
$ \Rightarrow {E_T} = 6.21 \times {10^{ - 21}}J/mol$
There are five degrees of freedom in an oxygen molecule. The total kinetic energy is given by the formula, ${E_T} = \dfrac{5}{2}kT$ so the total kinetic energy of an oxygen molecule at \[{27^0}C\]is; $ \Rightarrow {E_T} = \dfrac{5}{2} \times 1.38 \times {10^{ - 23}} \times 300$
$ \Rightarrow {E_T} = 10.35 \times {10^{ - 21}}J/mol$
The total kinetic energy in joule of one mole of oxygen at \[{27^0}C\] means we have to calculate its internal energy so the total kinetic energy in joule of one mole of oxygen at \[{27^0}C\] is given by the formula $U = \dfrac{f}{2}\mu RT$ where $U$is internal energy of molecule ,$f$ is degree of freedom of molecule ,$\mu $ is number of moles , $R$ is gas constant whose value is equals to $8.314J/mol - K$and $T$ is temperature . Now let's find out its internal energy , $U = \dfrac{5}{2} \times 1 \times 8.314 \times 300$ $ \Rightarrow U = 6.235.5J/mol$ .

Note: Here we have approached this problem by calculating the degree of freedom of molecules and we know that oxygen molecules have five degrees of freedom then with the help of some important formula we calculated its values.