Question

Given the Boltzmann’s constant, \[{k_B} = 1.38 \times {10^{ - 23}}{\text{ J/K}}\] , what will be the average translational kinetic energy of hydrogen gas molecules at NTP?
A. $0.186 \times {10^{ - 20}}{\text{ Joule}}$
B. $0.372 \times {10^{ - 20}}{\text{ Joule}}$
C. $0.56 \times {10^{ - 20}}{\text{ Joule}}$
D. $5.6 \times {10^{ - 20}}{\text{ Joule}}$

Answer 1

Hint: Average kinetic energy of gas molecules is given by $K.E. = \dfrac{3}{2}{k_B}T$ , where ${k_B}$ is Boltzmann’s constant having a value of $1.38 \times {10^{ - 23}}{\text{ J/K}}$ . This average kinetic energy is directly proportional to the absolute temperature of the gas molecules. .



Formula used:
Average kinetic energy of gas molecules is given by:
$K.E. = \dfrac{3}{2}{k_B}T$

Complete answer:
In the above question, we have to calculate the average Kinetic Energy of the gas molecules at NTP, that is, Normal temperature and pressure.
Thus, $T = 273{\text{ K}}$ .
Average kinetic energy of gas molecules is given by:
$K.E. = \dfrac{3}{2}{k_B}T$ … (1)
Here, $T$ is the absolute temperature and
${k_B}$ is Boltzmann’s constant having a value of $1.38 \times {10^{ - 23}}{\text{ J/K}}$ .
Substituting the value of temperature in equation (1) to calculate the value of the average translational kinetic energy,
$K.E. = \dfrac{3}{2}(1.38 \times {10^{ - 23}})(273) = 0.56 \times {10^{ - 20}}{\text{ J}}$
Hence, the average translational kinetic energy of hydrogen gas molecules at NYP is $0.56 \times {10^{ - 20}}{\text{ J}}$ .
Thus, the correct option is C.



Note: Average kinetic energy of gas molecules is given by $K.E. = \dfrac{3}{2}{k_B}T$ , where ${k_B}$ is Boltzmann’s constant. It is important to note that this average kinetic energy is dependent on temperature as it is directly proportional to it.