这是一份imperial帝国理工大学 F300作业代写的成功案例
$$
\begin{aligned}
\rho v+\Delta \rho c &=0, \
\mu c v+c_{p} \Delta T &=0 .
\end{aligned}
$$
$\Delta T$ can be related to $\Delta \rho$ through the adiabatic condition. Since $p^{1-\gamma} T^{\gamma}=$ constant and $\rho \propto \frac{p}{T}$ by the ideal gas law,
$$
\rho^{1-\gamma} T=c
$$
for some constant $c$. Taking the total derivative of the above,
$$
\begin{gathered}
(1-\gamma) \rho^{-\gamma} T d \rho+\rho^{1-\gamma} d T=0 \
d T=\frac{(\gamma-1) T}{\rho} d \rho .
\end{gathered}
$$
Since $\Delta \rho$ and $\Delta T$ are small,
$$
\Delta T \approx \frac{(\gamma-1) T}{\rho} \Delta \rho
$$
Substituting this expression for $\Delta T$ and summarizing our equations,
$$
\begin{aligned}
\mu c v+\frac{c_{p}(\gamma-1) T}{\rho} \Delta \rho &=0 \
\rho v+c \Delta \rho &=0
\end{aligned}
$$
F300 COURSE NOTES ：
$$
\begin{gathered}
\mu c^{2}-c_{p}(\gamma-1) T=0 \
c=\sqrt{\frac{c_{p}(\gamma-1) T}{\mu}} .
\end{gathered}
$$
Notice that $c_{p}(\gamma-1)=c_{p} \cdot \frac{c_{p}-c_{v}}{c_{v}}=\frac{c_{p}}{c_{v}} \cdot R=\gamma R$ and $\frac{\mu}{R T}=\frac{\rho}{p}$ by the ideal gas law such that the above becomes
$$
c=\sqrt{\frac{\gamma p}{\rho}}
$$