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
683.1k+ 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.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

