The translational kinetic energy of gas molec | vedclass

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

Question

(A) The translational kinetic energy per molecule of an ideal gas is given by $K.E._{molecule} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann const

Vedclass · Accepted Answer

$\frac{3}{2}RT$

Vedclass · Answer

(A) The translational kinetic energy per molecule of an ideal gas is given by $K.E._{molecule} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.For one mole of gas,the number of molecules is equal to Avogadro's number,$N_A$.Therefore,the total translational kinetic energy for one mole is $K.E._{molar} = N_A 	imes (\frac{3}{2} k_B T)$.Since the universal gas constant $R = N_A k_B$,we substitute this into the equation.Thus,$K.E._{molar} = \frac{3}{2} RT$.

The translational kinetic energy of gas molecules for one mole of the gas is equal to

Similar Questions

By using which equation can the force acting on the walls of a container due to gas molecules be obtained?

The temperature of a gas is a measure of

The mean kinetic energy of the molecules of an ideal gas at $399^{\circ} C$ is $E$. The temperature at which the mean kinetic energy of its molecules will be $E/2$ is: (in $^{\circ} C$)

At what temperature in $^\circ C$ is the average molecular kinetic energy of any gas double that of its value at $20^\circ C$?

The total kinetic energy of $1 \text{ mole}$ of $N_2$ at $27^{\circ}C$ is approximately: $(R = 2 \text{ cal/mol K})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module