At 300 K,the number of H_2 molecules is twic | vedclass

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

Question

(C) The total kinetic energy $(E)$ of an ideal gas is given by $E = N 	imes \frac{f}{2} kT$,where $N$ is the number of molecules,$f$ is the degrees o

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(C) The total kinetic energy $(E)$ of an ideal gas is given by $E = N 	imes \frac{f}{2} kT$,where $N$ is the number of molecules,$f$ is the degrees of freedom,$k$ is the Boltzmann constant,and $T$ is the temperature.For $H_2$ (a diatomic gas),the degrees of freedom $f = 5$ at $300 \ K$.For $O_2$ (a diatomic gas),the degrees of freedom $f = 5$ at $300 \ K$.Given $N_{H_2} = 2N_{O_2}$.The ratio of total kinetic energy is $\frac{E_{H_2}}{E_{O_2}} = \frac{N_{H_2} 	imes \frac{f}{2} kT}{N_{O_2} 	imes \frac{f}{2} kT} = \frac{N_{H_2}}{N_{O_2}}$.Substituting the values,$\frac{E_{H_2}}{E_{O_2}} = \frac{2N_{O_2}}{N_{O_2}} = 2$.Thus,the ratio is $2:1$.

At $300 \ K$,the number of $H_2$ molecules is twice the number of $O_2$ molecules. What is the ratio of the total kinetic energy of $H_2$ to that of $O_2$?

Similar Questions

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

The kinetic energy of $20 \, L$ of ${H_2}$ gas is $1.5 \times 10^5 \, J$. Find the pressure exerted on the container.

On the basis of the kinetic theory of gases,the mean kinetic energy $(K.E.)$ of $1\, mole$ per degree of freedom is:

Explain the kinetic interpretation of temperature.

Molecular motion shows itself as

Vedclass Test Series

Exam Paper Generator

Online Exam Module