Question bank — Speed of sound — a = √(γRT) — derivation

False. Raising $p$ at fixed $T$ raises $\rho$ in exact proportion, so $p/\rho =RT$ is unchanged and $a=\sqrt{\gamma RT}$ is unchanged — stiffness and inertia rise together and cancel.

True for an ideal gas of fixed composition, because $a=\sqrt{\gamma RT}$ and $\gamma ,R$ are constants of the gas — only $T$ varies.

False. Colder gas has slower thermal molecular motion, so $a=\sqrt{\gamma RT}$ is smaller; sound is slower in cold air (295 m/s at 217 K vs 340 m/s at 288 K).

True. Helium's molar mass is tiny ($4$ g/mol) so its specific $R=R_u/M$ is large; lighter, faster molecules pass the push along quicker — hence the squeaky-voice effect (see Ideal gas law and specific gas constant).

False. The gas law always holds, but the process constraint here is adiabatic (isentropic): compression is too fast for heat to conduct away, so entropy is what stays constant, not temperature (see Adiabatic vs isothermal processes).

True mathematically — but no real gas has $\gamma =1$; for diatomic air $\gamma =1.4$ raises $a$ by $\sqrt{1.4}\approx 1.18$, the Laplace correction that fixed Newton's 15% error.

False. It is the specific gas constant $R=R_u/M$ (per kilogram), $287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ for air. Using $8.314$ gives a nonsensical $\sim 30$ m/s.

False. Steps giving $a^2=dp/d\rho$ used only mass and momentum conservation. The adiabatic assumption enters afterward, to evaluate $dp/d\rho =\gamma p/\rho$.

The error is holding $T$ constant during the differentiation. The wave is adiabatic, so $T$ also changes as $\rho$ changes; the correct constraint is $p\propto {\rho }^{\gamma }$, giving $dp/d\rho =\gamma RT$.

You physically can't, but a Galilean frame change is legitimate: subtracting the constant wave speed $a$ from all velocities turns the unsteady problem into a steady one without changing any physics — it's a bookkeeping choice.

That term is a product of two infinitesimals, i.e. second-order small, and vanishes relative to the first-order terms as the disturbance $\to 0$. Keeping it would contradict the "infinitesimal pulse" definition of $a$.

They are identical — the ideal gas law $p/\rho =RT$ makes them equal. The first is the general result; the second just substitutes the equation of state.

Speed $a$ is defined for an infinitesimal disturbance and is amplitude-independent. Finite-amplitude waves steepen and can form shocks, but that's outside the linear-acoustics regime (see Normal shock waves).

Constant entropy (isentropic) is not the same as constant temperature. An adiabatic compression raises $T$ while keeping $s$ fixed; isothermal would require heat flow, which contradicts adiabatic.

In the wave frame the flow is steady, so mass and momentum balances become simple algebraic in–out statements instead of time-dependent PDEs.

Mass and momentum give the mechanical relation $a^2=dp/d\rho$; the value of that derivative depends on the thermodynamic path, and only the adiabatic path $p\propto {\rho }^{\gamma }$ injects $\gamma$.

Sound is passed molecule to molecule by collisions, so it can't propagate faster than the molecules carrying it; both scale with $\sqrt{T}$, so $a$ and mean thermal speed track together.

Higher altitude means lower $T$, hence lower $a=\sqrt{\gamma RT}$; with $a$ smaller, $M=V/a$ grows even though $V$ is unchanged (see Mach number and flow regimes).

For an infinitesimal, smooth disturbance there are no finite gradients to cause dissipation, so the adiabatic process is also reversible — combined they mean isentropic, which is exactly $(\partial p/\partial \rho {)}_s$.

$a$ tracks the ratio $p/\rho =RT$; scaling $p$ and $\rho$ equally leaves the ratio fixed, whereas raising $T$ raises the ratio directly.

$a=\sqrt{\gamma RT}\to 0$: with no thermal motion there are no molecular collisions to relay the disturbance, so sound cannot propagate.

No. A shock is a finite, irreversible discontinuity where entropy jumps; the isentropic assumption fails there, so the linear speed-of-sound relation does not describe it (see Normal shock waves).

Only while the gas is dense enough to behave as a continuum; when the mean free path rivals the wavelength, there are too few collisions and classical sound propagation breaks down.

$V=a$, so $M=1$; disturbances can no longer outrun the object upstream, marking the boundary where compressibility becomes dominant (see Compressibility and why M > 0.3 matters).

The isothermal one, $a=\sqrt{RT}$ (Newton's value). This is a limiting case that real audible sound never satisfies because oscillations are far too fast.

The monatomic gas, since larger $\gamma$ raises $a=\sqrt{\gamma RT}$; more of the compression energy goes into pressure rather than internal molecular modes.

No — $R=R_u/M$ depends only on molar mass, a property of the molecule, not on the thermodynamic state.