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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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	$IV$. $\frac{5}{3}$

Similar Questions

If $\operatorname{cosec} \theta = \frac{p+q}{p-q}$,then $\cot \left(\frac{\pi}{4} + \frac{\theta}{2}\right)$ is equal to

An electron is travelling along the $x$-direction. It encounters a magnetic field in the $y$-direction. Its subsequent motion will be

Select the correct statement regarding the following change:
$NO_2^+ \to NO_2 \to NO_2^-$

For a gas $\frac{R}{C_v} = 0.4$,where $R$ is the universal gas constant and $C_v$ is the molar specific heat at constant volume. The gas is made up of molecules which are

The perimeter of a $\triangle ABC$ is $6$ times the arithmetic mean of the sine of its angles. If its side $BC$ is of unit length,then $\angle A=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module