The total random kinetic energy of 1 ,g of he | vedclass

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

Question

(B) Helium $(He)$ is a monoatomic gas. The degrees of freedom $(f)$ for a monoatomic gas is $3$.The total random kinetic energy $(U)$ of an ideal gas

Vedclass · Accepted Answer

$311.25$

Vedclass · Answer

(B) Helium $(He)$ is a monoatomic gas. The degrees of freedom $(f)$ for a monoatomic gas is $3$.The total random kinetic energy $(U)$ of an ideal gas is given by the formula: $U = \frac{f}{2} nRT$,where $n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.Given: Mass $(m)$ = $1 \,g$,Molar mass of Helium $(M)$ = $4 \,g/mol$,Temperature $(T)$ = $100 \,K$,$R = 8.3 \,J \,mol^{-1} \,K^{-1}$.Number of moles $(n)$ = $\frac{m}{M} = \frac{1}{4} = 0.25 \,mol$.Substituting the values into the formula: $U = \frac{3}{2} 	imes 0.25 	imes 8.3 	imes 100$.$U = 1.5 	imes 0.25 	imes 830$.$U = 0.375 	imes 830 = 311.25 \,J$.Therefore,the total random kinetic energy is $311.25 \,J$.

The total random kinetic energy of $1 \,g$ of helium at $100 \,K$ will be $\left(R=8.3 \,J \,mol^{-1} \,K^{-1}\right)$. (in $\,J$)

Similar Questions

The kinetic energy of translation of $20 \, g$ of oxygen at $47^{\circ}C$ is (molecular weight of oxygen is $32 \, g/mol$ and $R = 8.3 \, J/mol \cdot K$).

In the case of hydrogen and oxygen at $N.T.P.$,which of the following quantities is/are the same?

At constant volume,the temperature of a gas is increased. Then:

The graph which represents the variation of mean kinetic energy of molecules with temperature $t^{\circ} \text{C}$ is

The translational kinetic energy of the molecules of a gas at absolute temperature $T$ can be doubled by

Vedclass Test Series

Exam Paper Generator

Online Exam Module