The total number of degrees of freedom in 1 | vedclass

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

Question

(D) At $NTP$,$22400 \ cm^3$ of any ideal gas contains $6.02 	imes 10^{23}$ molecules.Since $H_2$ is a diatomic molecule,the number of atoms per molec

Vedclass · Accepted Answer

$1.34375 	imes 10^{20}$

Vedclass · Answer

(D) At $NTP$,$22400 \ cm^3$ of any ideal gas contains $6.02 	imes 10^{23}$ molecules.Since $H_2$ is a diatomic molecule,the number of atoms per molecule is $2$. However,the question asks for the degrees of freedom of the molecules in $1 \ cm^3$.The number of molecules in $1 \ cm^3$ is $n = \frac{6.02 	imes 10^{23}}{22400} \approx 0.26875 	imes 10^{20}$ molecules.$A$ diatomic molecule $(H_2)$ has $5$ degrees of freedom ($3$ translational + $2$ rotational).Total degrees of freedom = (Number of molecules) $	imes$ (Degrees of freedom per molecule)Total degrees of freedom = $0.26875 	imes 10^{20} 	imes 5 = 1.34375 	imes 10^{20}$.

The total number of degrees of freedom in $1 \ cm^3$ of $H_2$ gas at $NTP$ is ....

Similar Questions

If $\gamma$ is the ratio of specific heats of an ideal gas,the number of degrees of freedom of the gas molecules is .......

The average thermal energy for a mono-atomic gas is: ($k_{B}$ is Boltzmann constant and $T$ is absolute temperature)

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature $T$? ($k_{B}$ is Boltzmann constant)

If the internal energy of $n_1$ moles of $He$ at temperature $10T$ is equal to the internal energy of $n_2$ moles of hydrogen $(H_2)$ at temperature $6T$,find the ratio $\frac{n_1}{n_2}$.

The internal energy of one mole of a rigid diatomic gas at absolute temperature $T$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module