Question

The mean kinetic energy of a gas molecule at $=$ ${27^0}C$ is $6.21 \times {10^{ - 21}}Joule$ . Its value at ${227^0}C$ will be:
A. $12.35 \times {10^{ - 21}}\,J$
B. $11.35 \times {10^{ - 21}}\,J$
C. $10.35 \times {10^{ - 21}}\,J$
D. $9.35 \times {10^{ - 21}}\,J$

Answer 1

In Kinetic theory of gases, we know that the mean energy of gas molecules is directly proportional to temperature. Here, we will use the general formula of the mean kinetic energy of a gas molecule and will solve for a given temperature.

Complete step by step answer:
The general formula for average kinetic energy in gas molecule is given by:
$KE = \dfrac{3}{2}KT$
where, $K = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$ which is known as Boltzmann constant and $T$ is the temperature of gas.

Given, we have $KE =$ $6.21 \times {10^{ - 21}}\,J$ at a temperature of $T = {27^0}C$. Now, we need to find the value of mean kinetic energy at a temperature of ${227^0}C$. Firstly, we will convert the given temperature into a Kelvin scale. The formula of conversion from celsius scale to kelvin scale is given as:
$273{ + ^0}C = kelvin$
$T = 227 + 273$
$T = 500K$
Hence, we now have temperature and the Boltzmann constant which has a value of $K = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$.
Put these values in formula of mean kinetic energy which is $KE = \dfrac{3}{2}KT$
$K.E = \dfrac{3}{2}KT$
$\Rightarrow K.E = 1.5 \times 1.38 \times {10^{ - 23}} \times 500$
$\therefore K.E = 10.35 \times {10^{ - 21}}\,J$
Hence, we find that the value of mean kinetic energy of gas molecules at a temperature of ${227^0}C$ is $K.E =$ $10.35 \times {10^{ - 21}}Joule$ .

Hence, the correct option is C.

Note: We should remember that, Boltzmann constant is the proportionality factor that releases the average relative kinetic energy of particles in a gas with the thermodynamic temperature of the gas. And the numerical value of Boltzmann constant is always constant which is $K = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$.