The ratio of C_p and C_v of a gas 'X' is 1.4. | vedclass

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(A) Given that the ratio $\frac{C_p}{C_v} = 1.4$,the gas '$X$' is diatomic.At $N.T.P.$,the volume of $1 	ext{ mole}$ of any ideal gas is $22.4 \ L$.N

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$6.02 	imes 10^{23}$

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(A) Given that the ratio $\frac{C_p}{C_v} = 1.4$,the gas '$X$' is diatomic.At $N.T.P.$,the volume of $1 	ext{ mole}$ of any ideal gas is $22.4 \ L$.Number of moles of gas '$X$' in $11.2 \ L$ = $\frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.Number of molecules = $	ext{moles} 	imes N_A = 0.5 	imes 6.022 	imes 10^{23} = 3.011 	imes 10^{23}$.Since the gas is diatomic,each molecule contains $2$ atoms.Number of atoms = $2 	imes (3.011 	imes 10^{23}) = 6.022 	imes 10^{23}$.

The ratio of $C_p$ and $C_v$ of a gas '$X$' is $1.4$. The number of atoms of the gas '$X$' present in $11.2 \ L$ of it at $N.T.P.$ is

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