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

If the difference between the specific heats of a gas is $4150 \, J/kg \cdot K$ and the ratio of specific heats is $1.4$,then the specific heat at constant volume is ...... $J/kg \cdot K$.

If $C_p$ and $C_v$ denote the specific heat of nitrogen at constant pressure and constant volume respectively,then

Obtain the equation for $\gamma = \frac{C_P}{C_V}$ in terms of the degree of freedom $f$.

If $C_p$ and $C_v$ denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively,then

One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} C$ to $100^{\circ} C$. The change in the internal energy of the gas is (Given, $R = 8.32 \text{ J mol}^{-1} \text{ K}^{-1}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module