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 $K.E.={\displaystyle \frac{3}{2}}{k}_{B}T$ , 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:
$K.E.={\displaystyle \frac{3}{2}}{k}_{B}T$

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:
$K.E.={\displaystyle \frac{3}{2}}{k}_{B}T$ … (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,
$K.E.={\displaystyle \frac{3}{2}}(1.38\times {10}^{-23})(273)=0.56\times {10}^{-20}\text{ J}$
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 $K.E.={\displaystyle \frac{3}{2}}{k}_{B}T$ , 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.