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

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(A) The ratio $\frac{C_p}{C_v} = 1.4$ indicates that the gas is diatomic (e.g.,$O_2, N_2$).Since the gas is diatomic,the number of atoms per molecule

Vedclass · Accepted Answer

$6.02 	imes 10^{23}$

Vedclass · Answer

(A) The ratio $\frac{C_p}{C_v} = 1.4$ indicates that the gas is diatomic (e.g.,$O_2, N_2$).Since the gas is diatomic,the number of atoms per molecule is $2$.At $NTP$,$1 \ mol$ of any gas occupies $22.4 \ L$.Therefore,moles of gas $'X' = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.Number of molecules $= 0.5 	imes N_A$.Number of atoms $= 0.5 	imes N_A 	imes 2 = N_A = 6.02 	imes 10^{23}$.

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 $NTP$ will be:

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