For a gas if gamma = 1.4,then atomicity,{C_p} | vedclass

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

Question

(C) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.Given $\gamma = 1.4$,we have $1.4 = 1 + \frac{2}{f}$

Vedclass · Accepted Answer

Diatomic,$\frac{7}{2}R, \frac{5}{2}R$

Vedclass · Answer

(C) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.Given $\gamma = 1.4$,we have $1.4 = 1 + \frac{2}{f}$.This implies $\frac{2}{f} = 0.4$,so $f = \frac{2}{0.4} = 5$.$A$ gas with $f = 5$ degrees of freedom is diatomic.The molar specific heat at constant volume is ${C_v} = \frac{f}{2}R = \frac{5}{2}R$.The molar specific heat at constant pressure is ${C_p} = {C_v} + R = \frac{5}{2}R + R = \frac{7}{2}R$.Therefore,the gas is diatomic,${C_p} = \frac{7}{2}R$,and ${C_v} = \frac{5}{2}R$.

For a gas if $\gamma = 1.4$,then atomicity,${C_p}$ and ${C_v}$ of the gas are respectively

Similar Questions

The specific heat at constant volume for the monoatomic argon is $0.075 \, kcal/kg-K$,whereas its gram molecular specific heat $C_V = 2.98 \, cal/mole/K$. The mass of the argon atom is (Avogadro's number $= 6.02 \times 10^{23} \, molecules/mole$)

For a gas,the difference between the two specific heats is $4150 \ J \ kg^{-1} \ K^{-1}$ and the ratio of the two specific heats is $1.4$. What is the specific heat of the gas at constant volume in units of $J \ kg^{-1} \ K^{-1}$?

The correct relation between the degree of freedom $f$ and the ratio of specific heat $\gamma$ is

The specific heat of argon at constant pressure and constant volume are $C_p$ and $C_v$ respectively. Its density $\rho$ at $N.T.P.$ will be: [Where $P$ and $T$ are pressure and temperature respectively at $N.T.P.$]

For an ideal gas of diatomic molecules,which of the following relations is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module