Question

For an ideal gas $\frac{Cp,m}{Cv,m}=\gamma$. The molecular mass of the gas is M , its specific heat capacity at constant volume is:

• A. $\frac{R}{M(\gamma -1)}$
• B. $\frac{M}{R(\gamma -1)}$
• C. $\frac{\gamma RM}{\gamma -1}$
• D. $\frac{\gamma R}{M(\gamma -1)}$

Answer 1

Answer: (A)

$\frac{R}{M(\gamma -1)}$

Solution:- (A) $\frac{R}{M(\gamma -1)}$

Let

As we know that,

$C_{p,m}-C_{v,m}=R.....\left(1\right)$

Dividing equation $\left(1\right)$ by $C_{v,m}$ we have

$\frac{C_{p,m}}{C_{v,m}}-\frac{C_{v,m}}{C_{v,m}}=\frac{R}{C_{v,m}}$

$\Rightarrow \frac{C_{p,m}}{C_{v,m}}-1=\frac{R}{C_{v,m}}$

Here,

$C_{p,m}=$ molar specific heat capacity at constant pressure

$C_{v,m}=$ molar specific heat capacity at constant volume

Given:-

$\frac{C_{p,m}}{C_{v,m}}=\gamma$

$\therefore \gamma -1=\frac{R}{C_{v,m}}$

As we know that,

$C_{v,m}=M{C}_v$

here, $C_v$ is specific heat capacity

$\therefore \gamma -1=\frac{R}{M{C}_v}$

$\Rightarrow C_V=\frac{R}{M(\gamma -1)}$

Hence the specific heat capacity of gas at constant volume is $\frac{R}{M(\gamma -1)}$.