If γ is the ratio of specific heats and R is the universal gas constant, then the molar specific heat at constant volume Cv is given by

Question

If &gamma; is the ratio of specific heats and R is the universal gas constant, then the molar specific heat at constant volume Cv is given by (A) &gamma; R (B)  ((&gamma;-1)R/&gamma;) (C)  (R/&gamma; -1) (D)  (&gamma; R/&gamma; -1)

Tardigrade Admin · Accepted Answer

From the Mayer's formula Cp-CV=R ...(i) and &gamma;=(Cp/CV) ⇒ &gamma; CV =Cp ...(ii) Substituting Eq. (ii) in Eq. (i), we get &gamma; CV-CV =R CV(&gamma;-1) =R CV =(R/&gamma;-1)

Q. If $γ$ 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

$γ R$

$\frac{( γ - 1 ) R}{γ}$

$\frac{R}{γ - 1}$

$\frac{γ R}{γ - 1}$

From the Mayer's formula
$C_{p} - C_{V} = R ... (i)$
and $γ = \frac{C _{p}}{C _{V}}$
$\Rightarrow γ C_{V} = C_{p} ... (ii)$
Substituting Eq. (ii) in Eq. (i), we get
$γ C_{V} - C_{V} = R$
$C_{V} (γ - 1) = R$
$C_{V} = \frac{R}{γ - 1}$

Q. If γ is the ratio of specific heats and R is the universal gas constant, then the molar specific heat at constant volume Cv​ is given by

γR

γ(γ−1)R​

γ−1R​

γ−1γR​