Question

As the temperature is raised from $ {20^o}C $ to $ {40^o}C $ , the average kinetic energy of neon atoms changes by a factor:
A. $ 2 $
B. $ \sqrt {\dfrac{{313}}{{293}}} $
C. $ \dfrac{{313}}{{293}} $
D. $ \dfrac{1}{2} $

Answer 1

Hint: The temperature changes can result in the specific changes in the kinetic energy. The velocity of the molecule changes which results in the shift in kinetic energy and for every gas this condition is applicable based on the characteristic changes in the physical property.

Complete step by step answer
The temperature gets raised for the given inert gas from $ {20^o}C $ to $ {40^o}C $ . The temperature is one of the physical factors which causes changes in other physical properties of the gases. One of the changes that occur due to changing temperature is the changes in the kinetic energy of the atoms of the given has. The kinetic energy of neon changes with a given temperature. The movement of the atoms of a particular element changes because as the temperature changes the volume of the gas changes with it. The volume of neon gas changes so that there is a higher intermolecular space for the atoms to move easily. This is why the kinetic energy changes as there is a higher space for the movement of the atoms.
The kinetic energy of the neon gases: $ KE = \dfrac{3}{2}RT $
The ratio of the kinetic energy at $ {20^o}C $ and at $ {40^o}C $ when given in a ratio will be calculated using the value of $ K{E_{40}} $ and $ K{E_{20}} $ . The ratio can be calculated as
 $ \dfrac{{K{E_{40}}}}{{K{E_{20}}}} = \dfrac{{\dfrac{3}{2}R{T_{40}}}}{{\dfrac{3}{2}R{T_{20}}}} $
 $ \Rightarrow \dfrac{{K{E_{40}}}}{{K{E_{20}}}} = \dfrac{{{T_{40}}}}{{{T_{20}}}} $ (Since the other components are constant and hence cancelled out)
Here the temperature is calculated in the form of Kelvin scale, which is why all the temperatures need to be converted to Kelvin before finding out the ratio.
 $ \dfrac{{K{E_{40}}}}{{K{E_{20}}}} = \dfrac{{273 + 40}}{{273 + 20}} $
 $ \Rightarrow \dfrac{{K{E_{40}}}}{{K{E_{20}}}} = \dfrac{{313}}{{293}} $
 $ \Rightarrow K{E_{40}} = \dfrac{{313}}{{293}}K{E_{20}} $
Therefore, the correct option about the times in which the kinetic energy changes is C. $ \dfrac{{313}}{{293}} $ .

Note
The kinetic energy of the atom can be calculated as each of the atoms move inside the container in which the gas is contained. The Brownian motion provides the velocity to the given atom and hence the kinetic energy of the atoms is generated in the system.