Match the frac{C_P}{C_V} ratio for ideal gase | vedclass

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Question

(A) The ratio $\gamma = \frac{C_P}{C_V}$ is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.$(A)$ Monoatomic: $f = 3$,$\gamma =

Vedclass · Accepted Answer

$A-IV, B-I, C-II, D-III$

Vedclass · Answer

(A) The ratio $\gamma = \frac{C_P}{C_V}$ is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.$(A)$ Monoatomic: $f = 3$,$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$ $(IV)$.$(B)$ Diatomic rigid molecules: $f = 5$,$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$ $(I)$.$(C)$ Diatomic non-rigid molecules: $f = 7$,$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$ $(II)$.$(D)$ Triatomic rigid molecules: $f = 6$,$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$ $(III)$.Thus,the correct matching is $A-IV, B-I, C-II, D-III$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

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