Question

Match the $\frac{C_p}{C_v}$ ratio for ideal gases with different types of molecules:

Column - I Molecule Type	Column -II $\frac{C_p}{C_v}$
$\left(A\right)$ Monatomic	$\left(I\right)\frac{7}{5}$
$\left(B\right)$ Diatomic rigid molecules	$\left(II\right)\frac{9}{7}$
$\left(C\right)$ Diatomic non-rigid molecules	$\left(III\right)\frac{4}{3}$
$\left(D\right)$ Triatomic rigid molecules	$\left(IV\right)\frac{5}{3}$

• A. $\text{(A)-(IV), (B)-(II), (C)-(I),(D)-(III)}$
• B. $\text{(A)-(III), (B)-(IV), (C)-(II),(D)-(I)}$
• C. $\text{(A)-(IV), (B)-(I), (C)-(II),(D)-(III)}$
• D. $\text{(A)-(II), (B)-(III), (C)-(I),(D)-(IV)}$

Answer 1

The correct option is C $\text{(A)-(IV), (B)-(I), (C)-(II),(D)-(III)}$

As we know, $C_p=\frac{fR}{2}+R;C_v=\frac{fR}{2}$

$\gamma =\frac{C_p}{C_v}=1+\frac{2}{f}$

Where $f=$ degree of freedom

$\left(A\right)$ Monoatomic, $f=3$
$\therefore \gamma =1+\frac{2}{3}=\frac{5}{3}$

$\left(B\right)$ Diatomic rigid molecules, $f=5$
$\therefore \gamma =1+\frac{2}{5}=\frac{7}{5}$

$\left(C\right)$ Diatomic non-rigid molecules, $f=7$
$\therefore \gamma =1+\frac{2}{7}=\frac{9}{7}$

$\left(D\right)$ Triatomic rigid molecules, $f=6$
$\therefore \gamma =1+\frac{2}{6}=\frac{4}{3}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.