The average kinetic energy per molecule of an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The average kinetic energy per molecule of an ideal gas is given by the formula $KE_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann consta

Vedclass · Accepted Answer

$6.17 	imes 10^{-21} \, J$

Vedclass · Answer

(B) The average kinetic energy per molecule of an ideal gas is given by the formula $KE_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant $(1.38 	imes 10^{-23} \, J/K)$ and $T$ is the temperature in Kelvin.Given $T = 25 + 273.15 = 298.15 \, K$.Substituting the values: $KE_{avg} = \frac{3}{2} 	imes (1.38 	imes 10^{-23} \, J/K) 	imes 298.15 \, K$.$KE_{avg} = 1.5 	imes 1.38 	imes 298.15 	imes 10^{-23} \, J$.$KE_{avg} \approx 6.17 	imes 10^{-21} \, J$.

The average kinetic energy per molecule of an ideal gas at $25\,^oC$ in $SI$ units is.....

Similar Questions

The kinetic energy of $1 \ mol$ of an ideal gas is given by:

The average kinetic energy of one molecule of an ideal gas at $27^{\circ} C$ and $1 \ atm$ pressure is

$A$ helium atom is two times heavier than a hydrogen molecule. At $298 \ K$,the average kinetic energy of helium is:

Which of the following relations is correct for an ideal gas?

$A$ football bladder contains an equimolar mixture of $H_2$ and $O_2$. What is the ratio of the rate of diffusion of $H_2$ to $O_2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module