In gases of diatomic molecules,the ratio of t | vedclass

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

Question

(B) For a diatomic gas,the number of degrees of freedom $f = 5$ ($3$ translational and $2$ rotational at room temperature).The molar specific heat at

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(B) For a diatomic gas,the number of degrees of freedom $f = 5$ ($3$ translational and $2$ rotational at room temperature).The molar specific heat at constant volume is given by $C_V = \frac{f}{2}R = \frac{5}{2}R$.The molar specific heat at constant pressure is given by $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R$.The ratio of specific heats $\gamma$ is defined as $\gamma = \frac{C_P}{C_V}$.Substituting the values,$\gamma = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.

In gases of diatomic molecules,the ratio of the two specific heats of gases ${C_P}/{C_V}$ is

Similar Questions

In a non-rigid diatomic molecule with an additional vibrational mode,what is the relationship between $C_v$ and $C_p$?

If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant,then the molar specific heat at constant volume $C_{V}$ is given by

To increase the temperature of $1$ mole of an ideal monatomic gas by $10^{\circ}C$ at constant pressure,$40 \, cal$ of heat is required. How much heat (in $cal$) is required to increase the temperature by the same amount at constant volume?

The amount of heat energy required to raise the temperature of $1\,g$ of helium from $T_1\,K$ to $T_2\,K$ is

What is the correct relationship between molar specific heat at constant pressure $(C_P)$ and constant volume $(C_V)$? ($R$ is the universal gas constant)

Vedclass Test Series

Exam Paper Generator

Online Exam Module