Isentropic Pressure Ratio

The pressure form of the same physics, raised to $\gamma/(\gamma-1)$: $p_0/p = \left(1 + \frac{\gamma-1}{2}M^{2}\right)^{\gamma/(\gamma-1)}$. That exponent is 3.5 for air and 6 for a $\gamma = 1.2$ exhaust, and it is why pressure ratios run away so much faster than temperature ratios. At Mach 3 in air the gas is 2.8 times hotter when brought to rest but 36.7 times higher in pressure.

Read backwards, this is how a nozzle is designed. Fix the chamber pressure, decide what exit pressure you want to match against the ambient you will be flying in, and the ratio hands you the exit Mach number — which the area relation then converts into an expansion ratio and a bell you can draw. A 7 MPa chamber expanded to Mach 3 at $\gamma = 1.2$ exits at 149 kPa, slightly above sea-level ambient, which is a mildly under-expanded sea-level nozzle. Want to exit at 5 kPa for high altitude and the required Mach number, and the bell, both grow substantially.

The same relation runs an air data computer. A pitot-static probe measures total and static pressure, their ratio gives Mach directly, and an aircraft therefore reads Mach without ever measuring the speed of sound. Above Mach 1 the simple form fails, and it fails for a specific reason: a bow shock stands ahead of the probe, the flow through that shock is not isentropic, and the probe reads the total pressure BEHIND it. The correction is the Rayleigh supersonic pitot formula, and using this equation instead gives a Mach number that is confidently wrong.

Two housekeeping points that cause real errors. Both pressures must be ABSOLUTE — a gauge reading is already referenced to the atmosphere and produces nonsense here. And unlike total temperature, total pressure is NOT conserved across a shock: it drops, and the drop is precisely the entropy the shock generates. That loss is what makes supersonic inlet design a discipline of its own, since every percent of total pressure recovered upstream shows up as thrust downstream.