Solution:

Sonic Velocity for Isothermal Process:

For isothermal process : $\frac{p}{\rho}=$ constant

Differentiating both sides, we get,

$$ \frac{\rho \cdot d p-p \cdot d \rho}{\rho^{2}}=0 \quad \text { or } \quad \frac{d p}{\rho}-\frac{p \cdot d \rho}{\rho^{2}}=0 \\ $$

$$or,$$

$$ \frac{d p}{\rho}=\frac{p \cdot d \rho}{\rho^{2}} \quad \text { or } \quad \frac{d p}{d \rho}=\frac{p}{\rho}=R T...(1) \\ $$

$ \left(\frac{p}{\rho}=R T \quad\right.$..equation of state $) \\ $

Substituting the value of $\frac{d p}{d \rho}$, we get,

$$ C=\sqrt{\frac{p}{\rho}}=\sqrt{R T}...(2) \\ $$

Sonic Velocity for Adiabatic Process:

For isentropic (reversible adiabatic) process : $\frac{p}{\rho^{\gamma}}=$ constant or,

$$ p \cdot \rho^{-\gamma}=\text { constant } \\ $$

Differentiating both sides, we have,

$$ p(-\gamma) \cdot \rho^{-\gamma-1} d \rho+\rho^{-\gamma} d p=0 \\ $$

Dividing both sides by $\rho^{-\gamma}$, we get,

$$ p \gamma \rho^{-1} d \rho+d p=0 \text { or } d p=p \gamma \rho^{-1} d \rho \\ $$

$$or,$$

$$ \frac{d p}{d \rho}=\frac{p}{\rho} \gamma=\gamma R T \\ $$

$$ \left(\because \frac{p}{\rho}=R T\right) \\ $$

Substituting the value of, $\frac{d p}{d \rho}$, we get,

$$ C=\sqrt{\gamma R T} \\ $$

The following points are worth noting :

• (i) The process is assumed to be adiabatic when minor disturbances are to be propagated through air ; due to very high velocity of disturbances/pressure waves appreciable heat transfer does not take place.

• (ii) For calculation of velocity of the sound/pressure waves, isothermal process is considered only when it is mentioned in the numerical problem (that the process is isothermal).

• When no process is mentioned in the problem, calculation are made assuming the process to be adiabatic.