As said, these are perfect gas laws, but air, albeit being close to this ideal, is not entirely perfect.
So in a lamination valve the enthalpy stays ALMOST constant, resulting in temperature loss of just one-two Centigrade degrees (at moderate flow), while if the expansion was fully adiabatic-reversible (isoethropic) the temperature loss could be in the order of 100 centigrade degrees, or more.
So, a small temperature loss of 1-2 degrees is quite close to isotherm expansion, compared with the enormous temperature change which would happen in a truly iso-enthropic expansion.
Regarding the perfect gas law reported above, that describes EQUILIBRIUM conditions, it does not describe a fast adiabatic expansion, where the law is instead of the type p*V^n=constant.
The exponent n depends on the degree of reversibility (iso-enthropicity). For a biatomic gas, as air, a perfectly reversible (isoenthropic) adiabatic expansion will give you n=gamma=1.41. Instead, for a perfectly isoenthalpic expansion, n=1.000.
The real world is always in between. But at moderate gas flow, n is "almost" 1, something as 1.001, or perhaps 1.002, resulting in a minimal temperature loss.
Instead, when the gas flow is very fast (and a Scubapro MK25 can provide a VERY FAST gas flow, possible triple than a membrane reg) the exponent of the adiabatic expansion goes up, probably around 1.004, or perhaps even 1.005 (but still well below the maximum theoretical possible value of 1.41).
After a little math, one can rewrite the adiabatic equation in terms of p and T, instead of p and V, obtaining:
p*T^(n/(1-n)) = constant, which allows to compute the temperature after expansion, T2, knowing the temperature before expansion, T1:
T2 = T1 * (p1/p2)^((1-n)/n)
Assuming T=274 K, p1=200 bar, p2 = 8 bar, n=1.41 (isoenthropic adiabatic expansion), you get T2 = 107.5 K = -165.5 °C, that is way belong the freezing point!
Instead, with an almost-isoenthalpic expansion (which is what really occurs in your reg while breathing normally), the exponent is n=1.001, hence you get T2 = 273.1 K, which is still 0.1°C above the freezing point.
It is obvious that when you ask more air from the reg, you will get an expansion which is still almost isoenthalpic, but with an exponent which becomes, say 1.005. With such value you get T2=269.6 K = -3.4 °C, and you get freezing.
More info here:
Adiabatic process - Wikipedia
See the section titled "
Adiabatic free expansion of a gas", where you can read:
"Since this process does not involve any heat transfer or work, the first law of thermodynamics then implies that the net internal energy change of the system is zero.
For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible."