Question

The average kinetic energy of an ideal gas per molecule in SI units at ${25^ \circ }C$ will be:
A.$6.17 \times {10^{ - 21}}kJ$
B.$6.17 \times {10^{ - 21}}J$
C.$6.17 \times {10^{ - 20}}kJ$
D.$7.16 \times {10^{ - 20}}J$

Answer 1

Hint: We can calculate the average kinetic energy of an ideal gas with the help of Boltzmann constant and absolute temperature. The formula to calculate the average kinetic energy is,
$AverageK.E. = \dfrac{3}{2}kT$
Boltzmann constant is given as $k$.
The temperature is represented as $T$.

Complete step by step answer:
Given data contains,
Temperature is ${25^ \circ }C$.
We have to convert the value of degree Celsius to Kelvin. We can use the formula below to calculate kelvin from degree Celsius.
$T{ = ^ \circ }C + 273$
Let us now substitute the value of degree Celsius in the expression.
$T{ = ^ \circ }C + 273$
$T = 25 + 273$
On adding we get,
$T = 298K$
The temperature in Kelvin is $298K$.
We can calculate the average kinetic energy of an ideal gas with the help of Boltzmann constant and absolute temperature.
The formula to calculate the average kinetic energy of an ideal gas is,
$AverageK.E. = \dfrac{3}{2}kT$
Boltzmann constant is given as $k$.
The temperature is represented as $T$.
We can substitute the value of Boltzmann constant and temperature in the expression. The value of Boltzmann constant is $1.36 \times {10^{ - 23}}J/K$.
$AverageK.E. = \dfrac{3}{2}kT$
$AverageK.E. = \dfrac{3}{2}\left( {1.38 \times {{10}^{ - 23}}J/K} \right)\left( {298K} \right)$
$AverageK.E. = 6.17 \times {10^{ - 21}}J$
The average kinetic energy of an ideal gas per molecule in SI units at ${25^ \circ }C$ is $6.17 \times {10^{ - 21}}J$.
Therefore, the option (B) is correct.

Note:
An alternate method to calculate the average kinetic energy of an ideal gas per molecule in SI units at ${25^ \circ }C$ is given below,
The formula to calculate the average kinetic energy is,
$AverageK.E. = \dfrac{3}{2}kT$
Boltzmann constant is given as $k$.
The temperature is represented as $T$.
The formula is simplified as,
$AverageK.E. = \dfrac{3}{2}\dfrac{R}{N}T$
Here R is gas constant $\left( {8.313\,J/mol/K} \right)$ and N is Avogadro number $\left( {6.023 \times {{10}^{23}}mol} \right)$.
We have to convert the value of degree Celsius to Kelvin. We can use the formula below to calculate kelvin from degree Celsius.
$T{ = ^ \circ }C + 273$
Let us now substitute the value of degree Celsius in the expression.
$T{ = ^ \circ }C + 273$
$T = 25 + 273$
$T = 298K$
The temperature in Kelvin is $298K$.
Let us now substitute the value of gas constant and Avogadro number in the expression to calculate the average kinetic energy.
$AverageK.E. = \dfrac{3}{2}\dfrac{R}{N}T$
$AverageK.E. = \dfrac{3}{2} \times \dfrac{{\left( {8.313J/mol/K} \right)}}{{6.023 \times {{10}^{23}}mol}}\left( {298K} \right)$
$AverageK.E. = 6.17 \times {10^{ - 21}}J$
The average kinetic energy of an ideal gas per molecule in SI units at ${25^ \circ }C$ is $6.17 \times {10^{ - 21}}J$. Therefore, the option (B) is correct.