Question

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

Answer 1

Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and obeys the gas laws. While kinetic energy is the energy that it possesses due to its motion.

Complete step by step answer:
We know that the average kinetic energy of a gas is given by,
$K.E. = \dfrac{1}{2}fRT$
Where
$f$ = Transnational degree of freedom
R = gas constant
T = temperature
In order to find the average kinetic energy per molecule we have to divide the total kinetic energy by ${N_A}$ which is Rydberg's constant to specify the number of molecules.
Therefore average kinetic energy of an ideal gas per molecule is given by

$K.E. = \dfrac{1}{{{N_A}}}(\dfrac{1}{2}fRT)$
We know that,
Transnational Degree of freedom of monoatomic molecule is 3 given by
$f = 3$
Also we know that,
$R = k{N_A}$
Here, k is boltzmann constant given by,
$k = 1.36 \times {10^{( - 23)}}J/K$
On simplifying the formula we have
$K.E. = \dfrac{3}{2}kT$
At room temperature,
$T = 25 + 273 = 298K$

On substituting the values we get,

$K.E. = \dfrac{3}{2} \times 1.36 \times {10^{( - 23)}}J/K \times 298K$

On simplifying we get
$K.E = 6.17 \times {10^{( - 21)}}J$
The average kinetic energy of an ideal gas per molecule in SI units at 25℃ will be $6.17 \times {10^{( - 21)}}J$

So, the correct answer is Option B.

Note:
Temperature is a measure of the average kinetic energy of all the molecules in a gas. As the temperature and, therefore, kinetic energy, of a gas changes, the RMS speed of the gas molecules also changes. The RMS speed of the molecules is the square root of the average of each individual velocity squared.