According to the kinetic theory of gases,the | vedclass

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

Question

(D) According to the kinetic theory of gases,the average kinetic energy per molecule of an ideal gas is given by $E_{avg} = \frac{3}{2}kT$,where $k$ i

Vedclass · Accepted Answer

$\frac{3RT}{2}$

Vedclass · Answer

(D) According to the kinetic theory of gases,the average kinetic energy per molecule of an ideal gas is given by $E_{avg} = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.For $1$ mole of gas,the number of molecules is equal to Avogadro's number,$N_A$.Therefore,the total kinetic energy for $1$ mole is $E = N_A 	imes (\frac{3}{2}kT)$.Since $k = \frac{R}{N_A}$,we have $E = N_A 	imes \frac{3}{2} 	imes (\frac{R}{N_A}) 	imes T$.Simplifying this,we get $E = \frac{3}{2}RT$.Thus,the average kinetic energy of $1$ gram-mole (which is $1$ mole) of a gas is $\frac{3RT}{2}$.

According to the kinetic theory of gases,the average kinetic energy of $1$ gram-mole of a gas is given by which of the following expressions?

Similar Questions

The average translational kinetic energy of a molecule in a gas becomes equal to $0.69 \ eV$ at a temperature of about,[Boltzmann's constant $= 1.38 \times 10^{-23} \ J \ K^{-1}$] (in $^{\circ} C$)

$A$ cylinder has a capacity of $20 \, L$ and is filled with $H_2$ gas. The total average kinetic energy of the gas is $1.5 \times 10^5 \, J$. What is the pressure of the hydrogen gas?

We sit in a room with windows open. Then,

The kinetic energy of one gm-mole of a gas at normal temperature and pressure is $(R = 8.31 \text{ J/mol-K})$.

If $10^{22}$ gas molecules,each of mass $10^{-26} \ kg$,collide with a surface (perpendicular to it) elastically per second over an area of $1 \ m^2$ with a speed of $10^4 \ m/s$,the pressure exerted by the gas molecules will be of the order of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module