Yes, and you can measure whether you have adiabatic or isothermal (or something else) with the propagation of sound waves by measuring the speed of sound, which depends on the kind of the thermodynamic state. It turns out that the correct speed under usual conditions (at least for sound waves in air) indeed the assumption of adiabatic processes (i.e., the assumption of a perfect fluid) gives the correct speed of sound as measured:
https://en.wikipedia.org/wiki/Acoustic_wave
The adiabatic change of state can be derived easily as follows. Just take the 1st+2nd Law in the form
$$\mathrm{d} U=T \mathrm{d} S - p \mathrm{d} V.$$
On the other hand for an ideal gas
$$\mathrm{d} U=C_V \mathrm{d} T.$$
Since adiabatic change means ##\mathrm{d} Q = T \mathrm{d} S## this implies
$$C_V \mathrm{d} T = -p \mathrm{d} V.$$
Further you have the ideal-gas equation of state
$$p V=\nu R T$$
and thus
$$C_V \mathrm{d} T = -\nu R T \frac{\mathrm{d} V}{V}$$
or separating the variables
$$\frac{\mathrm{d} T}{T} = -\frac{\nu R}{C_V} \frac{\mathrm{d} V}{V}.$$
Integrating between two states leads to
$$\ln \left (\frac{T_2}{T_1} \right)=-\frac{\nu R}{C_V} \ln \left (\frac{V_2}{V_1} \right).$$
Using again the ideal-gas equation in the form
$$T=\frac{p V}{\nu R} \; \Rightarrow \; \frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}$$
leads to
$$\ln \left (\frac{p_2}{p_1} \right) =-\left (\frac{\nu R}{C_V}+1 \right) \ln \left (\frac{V_2}{V_1} \right)=-\gamma \ln \left (\frac{V_2}{V_1} \right)$$
Here the adiabatic coefficient is
$$\gamma=\frac{\nu R}{C_V}+1=\frac{C_V+\nu R}{C_V}=\frac{C_P}{C_V}.$$
using ##C_P=C_V+\nu R## for an ideal gas. Since air is mostly a gas with molecules with 2 atoms this gives ##\gamma=1.4##. With this the adiabatic change of state is determined by the adiabatic equation of state for the ideal gas,
$$p_2 V_2^{\gamma}=p_1 V_1^{\gamma}.$$
The speed of sound ##c_{\text{s}}## is given by the equation
$$c_{\text{s}}^2=\frac{\mathrm{d} p}{\mathrm{d} \rho}.$$
Since ##\rho=m/V## (with ##m## the constant mass of the gas) we can rewrite the adiabatic equation of state as
$$p_2 \rho_2^{-\gamma} = p_1 \rho_1^{-\gamma}.$$
Now setting ##\rho_1=\rho_0## and ##p_1=p_0## the air pressure and ##p_2=p## and ##\rho_2=\rho## the pressure within the sound wave (which differs only very little compared to the air pressure), you get
$$p=p_0 \left (\frac{\rho}{\rho_0} \right)^{\gamma} \; \Rightarrow \; c_2 = \frac{p_0}{\rho_0^{\gamma}} \gamma \rho^{\gamma-1} \simeq \gamma \frac{p_0}{\rho_0}.$$
In the last step we have used that ##p \simeq p_0##.
To compare this with the isothermal speed of sound, we can simply use the usual gas equation of state, which simplifies for ##T=\text{const}## to the Boyle-Mariott-Law
$$p_0 V_0 = p V \; \Rightarrow \; p=p_0 \frac{V_0}{V}=p_0 \frac{\rho}{\rho_0} \; \Rightarrow \; c_{\text{s}}^{(\text{iso-thermal})}=\frac{p_0}{\rho_0},$$
which was assumed by Newton assuming isothermal sound waves.
As one can see the adiabatic speed of sound is larger by a factor ##\sqrt{\gamma}##, which for air is ##\sqrt{1.4} \simeq 1.18##.