The total number of degrees of freedom in 1 | vedclass

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(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 ....

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