The molar specific heat of an ideal gas at constant pressure and constant volume is 'Cp' and 'Cv' respectively. If 'R' is the universal gas constant and the ratio of 'Cp' to 'Cv' is 'γ' then Cv =

Question

The molar specific heat of an ideal gas at constant pressure and constant volume is 'Cp' and 'Cv' respectively. If 'R' is the universal gas constant and the ratio of 'Cp' to 'Cv' is '&gamma;' then Cv = (A) (1 - &gamma;/1 + &gamma;) (B) (1 + &gamma;/1 - &gamma;) (C) ( &gamma; - 1 /R) (D) (R/ &gamma; - 1 )

Tardigrade Admin · Accepted Answer

According to Mayer formula, Cp-CV=R dots(i) where, Cp= specific heat at constant pressure, CV= specific heat at constant volume and R= gas constant Now, &gamma;=(Cp/CV) ⇒ Cp=&gamma; GV ...(ii) From Eqs. (i) and (ii), we get ⇒ &gamma; CV-CV=R ⇒ CV(&gamma;-1)=R ⇒ CV=(R/(&gamma;-1))

Q. The molar specific heat of an ideal gas at constant pressure and constant volume is $^{'} C_{p}^{'}$ and $^{'} C_{v}^{'}$ respectively. If 'R' is the universal gas constant and the ratio of $^{'} C_{p}^{'}$ to $^{'} C_{v}^{'}$ is $^{'} γ^{'}$ then $C_{v} =$

$\frac{1 - γ}{1 + γ}$

$\frac{1 + γ}{1 - γ}$

$\frac{γ - 1}{R}$

$\frac{R}{γ - 1}$

Q. The molar specific heat of an ideal gas at constant pressure and constant volume is ′Cp′​ and ′Cv′​ respectively. If 'R' is the universal gas constant and the ratio of ′Cp′​ to ′Cv′​ is ′γ′ then Cv​=

1+γ1−γ​

1−γ1+γ​

Rγ−1​

γ−1R​

According to Mayer formula, Cp​−CV​=R…(i) where, Cp​= specific heat at constant pressure, CV​= specific heat at constant volume and R= gas constant Now,γ=CV​Cp​​ ⇒Cp​=γGV​...(ii) From Eqs. (i) and (ii), we get ⇒γCV​−CV​=R⇒CV​(γ−1)=R ⇒CV​=(γ−1)R​

Q. The molar specific heat of an ideal gas at constant pressure and constant volume is $^{'} C_{p}^{'}$ and $^{'} C_{v}^{'}$ respectively. If 'R' is the universal gas constant and the ratio of $^{'} C_{p}^{'}$ to $^{'} C_{v}^{'}$ is $^{'} γ^{'}$ then $C_{v} =$

$\frac{1 - γ}{1 + γ}$

$\frac{1 + γ}{1 - γ}$

$\frac{γ - 1}{R}$

$\frac{R}{γ - 1}$