Question

For an ideal gas having molar mass $M$, specific heat at constant pressure can be given as:

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

Answer 1

Answer: (A)

$\frac{\gamma RM}{\gamma -1}$

$R=\frac{\overline{R}}{M}$

where,

$\overline{R}$= Universal gas constant

$R$= specific gas constant and

$\overline{R}=RM$

$\because C_P-C_V=\overline{R}$ & $\frac{C_P}{C_V}=\gamma$

$\therefore C_V=\frac{C_P}{\gamma }$

$\therefore C_P-\frac{C_P}{\gamma }=\overline{R}$

$\therefore \left(\frac{\gamma -1}{\gamma }\right)C_P=\overline{R}=RM$

$\therefore C_P=\frac{\gamma RM}{\gamma -1}$

Hence, option (A) is the correct answer.