The average kinetic energy of an ideal gas per molecule in SI units at ${25^0}C$ will be
\[
A.{\text{ }}6.17 \times {10^{ - 21}}J{K^{ - 1}} \\
B.{\text{ }}6.17 \times {10^{ - 21}}kJ{K^{ - 1}} \\
C.{\text{ }}6.17 \times {10^{20}}J{K^{ - 1}} \\
D.{\text{ }}7.16 \times {10^{ - 21}}J{K^{ - 1}} \\
\]
Answer
664.8k+ views
Hint- In order to deal with this question we will use the formula of average kinetic energy of an ideal gas per molecule because in this question absolute temperature is given and we also know the value of Boltzmann constant so by putting these values we will get the answer easily.
Formula used- ${\text{Average kinetic energy}} = \dfrac{3}{2}kT$
Complete step-by-step answer:
Given that:
Absolute temperature $ = {25^0}C$
We know that the average kinetic energy of an ideal gas per molecule is given by the expression
Average kinetic energy $ = \dfrac{3}{2}kT$
Here, $k$ is Boltzmann constant and $T$ is absolute temperature.
$k = 1.36 \times {10^{ - 23}}J/K$
And now let us convert the temperature given from Celsius to Kelvin for using the same in the given formula
$
T = {25^0}C \\
T = 25 + 273 = 298K \\
$
Substitute the values of $k$ and $T$ in above formula we have
Average kinetic energy:
$
= \dfrac{3}{2}kT \\
= \dfrac{3}{2} \times 1.36 \times {10^{ - 23}}J/K \times 298K \\
= 6.17 \times {10^{ - 21}}J \\
$
Hence, the average kinetic energy of an ideal gas per molecule is $6.17 \times {10^{ - 21}}J$
So, the correct answer is option A.
Note- The Boltzmann constant is the proportionality factor that relates the average relative kinetic energy of particles in a gas with the thermodynamic temperature of the gas. It occurs in the definitions of the Kelvin and the gas constant, and in Planck's law of black-body radiation and Boltzmann's entropy formula. An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to inter particle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.
Formula used- ${\text{Average kinetic energy}} = \dfrac{3}{2}kT$
Complete step-by-step answer:
Given that:
Absolute temperature $ = {25^0}C$
We know that the average kinetic energy of an ideal gas per molecule is given by the expression
Average kinetic energy $ = \dfrac{3}{2}kT$
Here, $k$ is Boltzmann constant and $T$ is absolute temperature.
$k = 1.36 \times {10^{ - 23}}J/K$
And now let us convert the temperature given from Celsius to Kelvin for using the same in the given formula
$
T = {25^0}C \\
T = 25 + 273 = 298K \\
$
Substitute the values of $k$ and $T$ in above formula we have
Average kinetic energy:
$
= \dfrac{3}{2}kT \\
= \dfrac{3}{2} \times 1.36 \times {10^{ - 23}}J/K \times 298K \\
= 6.17 \times {10^{ - 21}}J \\
$
Hence, the average kinetic energy of an ideal gas per molecule is $6.17 \times {10^{ - 21}}J$
So, the correct answer is option A.
Note- The Boltzmann constant is the proportionality factor that relates the average relative kinetic energy of particles in a gas with the thermodynamic temperature of the gas. It occurs in the definitions of the Kelvin and the gas constant, and in Planck's law of black-body radiation and Boltzmann's entropy formula. An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to inter particle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.
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