Question

The average kinetic energy of a gas at $ - {23^0}C\,and\,75cm$ pressure is $5 \times {10^{ - 14}}erg$ for ${H_2}$. The mean kinetic energy of the ${O_{2\,}}\,at\,{227^0}C$ and $150cm$ pressure will be

(A) $40 \times {10^{ - 14}}erg$
(B) $10 \times {10^{ - 14}}erg$
(C) $20 \times {10^{ - 14}}erg$
(D) $80 \times {10^{ - 14}}erg$

Answer 1

Hint:First start with finding the relation between the average kinetic energy and the temperature of the gas. After finding the relation, use the information provided in the question such as kinetic energy of Hydrogen, temperature of Hydrogen and Oxygen gasses and get the required answer that is kinetic energy of Oxygen gas.


Formula used:
The average kinetic energy is
$K = \dfrac{1}{2}m \times \dfrac{{3RT}}{M}$


Complete answer:
Now, the formula for the average kinetic energy is as follows;
$K = \dfrac{1}{2}mV_{rms}^2$
Where, K is average kinetic energy
m is mass
${V_{rms}}$ is root mean square velocity
Putting value of root mean square velocity:
$K = \dfrac{1}{2}m \times \dfrac{{3RT}}{M}$
Since all the values are constant, we get;
$\dfrac{K}{T} = $ constant
So, $\dfrac{{{K_{{H_2}}}}}{{{K_{{O_2}}}}} = \dfrac{{{T_{{H_2}}}}}{{{T_{{O_2}}}}}$
${K_{{O_2}}} = \dfrac{{{K_{{H_2}}} \times {T_{{O_2}}}}}{{{T_{{H_2}}}}}$
${K_{{O_2}}} = \dfrac{{5 \times {{10}^{ - 14}} \times 500}}{{250}} = 10 \times {10^{ - 14}}erg$

Hence the correct answer is Option(B).



Note: Use the formula for the average kinetic energy carefully, know what are the constant values and what are the variable values and use accordingly. Be careful about the unit of all the quantities; it should be the same while putting in the formula in order to get the correct answer for the given question.