# Ideal Exhaust velocity of a rocket nozzle

In a youtube video, they defined exit exhaust velocity of a rocket nozzle as

$$v_2 = \sqrt{\dfrac{2k}{k - 1}RT_1 \left[1 - \left(\dfrac{p_2}{p_1}\right)^{(k - 1) / k}\right]}$$

How can I derive this equation? I am thinking this equation is derived from a lot of substitution using isentropic flow equations, but I am not sure. I've been stuck on this for a long time so help would be much appreciated!

First, let's identify the symbols: $$T_1$$ and $$p_1$$ are the combustion chamber temperature and pressure; $$k$$ is the ratio of specific heats and $$R$$ is gas constant for the combustion products; $$p_2$$ is the exit pressure.
There are many assumptions to simplify the calculations used to derive the exhaust gas equation. The combustion chamber is basically the stagnation condition and the enthalpy does not change in the expansion of the gases so $$h_c = h_e + \frac{1}{2}V_e^2$$
By definition, $$k = c_p /c_v$$ and for perfect gas $$c_p - c_v =R$$ . Hint, write $$c_p$$ as function of k and R. You will see it in your equation. Take enthalpy as $$c_p T$$. As the flow expands from the combustion chamber to the exit the velocity increases, the temperature decreases and pressure decreases also. We assume isentropic flow so $$T \propto p^{\frac{k-1}{k}}$$.
One of my old books used Rocketdyne as the source with the following values: $$T_c=5570 F, k=1.25, M_c = 21, I_{sp}= 266$$ for RP-1 and Liquid oxygen from 500 psi to 14.7 psi. Converting to SI, $$T_c = 3350 K$$, $$I_{sp}$$ converts to $$v_e = 2609 m/s$$. Using $$R_{universal}=8.3145 J/mole K$$, then $$R = 8.3145/.021kg/mole= 395.92 m^2/s^2 K$$.
I calculated $$v_e = 2591 m/s$$, so within 1% of the book value. Unfortunately most tables of propellants and specific impulse do not give $$M_c$$ and $$k$$.