Exercises — Speed of sound — a = √(γRT) — derivation

Temperature only. From $a=\sqrt{\gamma RT}$, the only variable is $T$; $\gamma$ and $R$ are fixed for a given gas. Although $p$ and $\rho$ appear in the form $\sqrt{\gamma p/\rho }$, their ratio $p/\rho =RT$ is set entirely by $T$. See Ideal gas law and specific gas constant.

Adiabatic (in fact isentropic). The oscillation is far too fast for heat to conduct across the wave in one cycle. The symbol that records this choice is $\gamma =c_p/c_v$. If the process were isothermal we would get $a=\sqrt{RT}$ with no $\gamma$. See Adiabatic vs isothermal processes.

$a^2={\left(\frac{\partial p}{\partial \rho }\right)}_s$ The subscript $s$ means "at constant entropy" — the derivative is taken along an isentropic (reversible adiabatic) path. See Isentropic relations p ∝ ρ^γ.

$a=\sqrt{\gamma RT}=\sqrt{1.4\times 287\times 300}=\sqrt{120,540}\approx 347m/s$ What we did: plugged $T$ straight into $\sqrt{\gamma RT}$. Why: we were handed a temperature, so this is the face to use.

First convert: $T=-50+273=223\mathrm{K}$ (never use $-50$ directly — the square root of a negative is meaningless here). $a=\sqrt{1.4\times 287\times 223}=\sqrt{89,597}\approx 299m/s$ Why lower than sea level: colder air → slower molecules → slower sound.

$M=\frac{V}{a}=\frac{272}{340}=0.80$ Since $0.3<M<0.8$, this is high subsonic and compressibility matters ($M>0.3$). See Mach number and flow regimes and Compressibility and why M > 0.3 matters.

Way 1 (given $p,\rho$ directly): $a=\sqrt{\gamma \frac{p}{\rho }}=\sqrt{1.4\times \frac{60,000}{0.80}}=\sqrt{1.4\times 75,000}=\sqrt{105,000}\approx 324m/s$ Way 2 (get $T$ first): from $p=\rho RT$, $T=\frac{p}{\rho R}=\frac{60,000}{0.80\times 287}=261.3\mathrm{K},a=\sqrt{1.4\times 287\times 261.3}\approx 324m/s$ They match, because $\gamma p/\rho =\gamma RT$ is the same number written two ways. Why do both: it proves you understand that $p/\rho$ is $RT$, the heart of "$a$ depends on $T$ only."

Sea level: $a=\sqrt{1.4\times 287\times 288}=\sqrt{115,718}\approx 340m/s$, so $M=250/340\approx 0.735$. Altitude: $a=\sqrt{1.4\times 287\times 217}=\sqrt{87,175}\approx 295m/s$, so $M=250/295\approx 0.847$. The altitude aircraft has the higher Mach number. Same speed, but the colder air carries sound more slowly, so the same $V$ is a bigger fraction of $a$. This is exactly why cruising jets watch their Mach limit, not their true airspeed.

First the specific constant: $R_{\text{He}}=R_u/M=8314/4=2078.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ (using $R_u=8314\mathrm{J}\mathrm{kmo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $M=4g/mol=4kg/kmol$). $a=\sqrt{1.667\times 2078.5\times 288}=\sqrt{997,860}\approx 999m/s$ That is about $999/340\approx 2.9\times$ the speed in air. Why: helium molecules are far lighter (small $M$ → large $R$), so at the same temperature they zip around faster and hand the pressure pulse along more quickly. This is the physics behind the squeaky-voice effect.

Newton (isothermal): $a_N=\sqrt{RT}=\sqrt{287\times 288}=\sqrt{82,656}\approx 287.5m/s$. Correct (adiabatic): $a=\sqrt{\gamma RT}=\sqrt{1.4}\times a_N=1.1832\times 287.5\approx 340m/s$. Error: $\frac{a_N-a}{a}=\frac{287.5-340.2}{340.2}\approx -0.155=-15.5\%$ Newton was about $15.5\%$ too low. The whole discrepancy is the missing factor $\sqrt{\gamma }=\sqrt{1.4}$. Why the fix works: the real compression traps its heat (adiabatic), making the gas stiffer than the isothermal model assumes — a stiffer medium carries sound faster. See Adiabatic vs isothermal processes.

Scaling: for one gas $\gamma R$ is constant, so $a=\sqrt{\gamma R}\sqrt{T}\Rightarrow a\propto \sqrt{T}$. Therefore $\frac{a_2}{a_1}=\sqrt{\frac{T_2}{T_1}}.$ Apply: $a_2=340\times \sqrt{\frac{350}{288}}=340\times \sqrt{1.2153}=340\times 1.1024\approx 375m/s$ Why this is powerful: the ratio trick cancels $\gamma$ and $R$ entirely, so it works for any gas as long as it stays the same gas. A $22\%$ rise in $T$ gives only an $\sqrt{1.22}-1\approx 10\%$ rise in $a$ — the square root tames it.

$V=Ma=M\sqrt{\gamma RT}$. Sea level: $a_1=\sqrt{1.4\times 287\times 288}\approx 340.2m/s$, so $V_1=0.85\times 340.2\approx 289m/s$. Altitude: $a_2=\sqrt{1.4\times 287\times 223}\approx 299.3m/s$, so $V_2=0.85\times 299.3\approx 254m/s$. Comment: to keep the same Mach number in colder, thinner air the aircraft actually flies slower in true airspeed (from $289$ to $254m/s$), because $a$ has dropped. Mach number, not raw speed, is what governs the shock/compressibility behaviour. See Mach number and flow regimes.

Step A — differentiate the isentropic law. With $C$ constant, $\frac{dp}{d\rho }=C\gamma {\rho }^{\gamma -1}=\gamma \frac{C{\rho }^{\gamma }}{\rho }=\gamma \frac{p}{\rho }.$ Why: $a^2=dp/d\rho$ came from mass and momentum alone; now we supply the process ($p\propto {\rho }^{\gamma }$) that tells us how $p$ actually moves with $\rho$. Step B — insert into $a^2$. $a^2=\frac{dp}{d\rho }=\gamma \frac{p}{\rho }$. Step C — use the ideal gas law $p=\rho RT\Rightarrow p/\rho =RT$: $a^2=\gamma RT\Rightarrow a=\sqrt{\gamma RT}.$ Evaluate at $250\mathrm{K}$: $a=\sqrt{1.4\times 287\times 250}=\sqrt{100,450}\approx 317m/s$.

Upstream: $a_1=\sqrt{1.4\times 287\times 250}=\sqrt{100,450}\approx 316.9m/s$. Downstream: $a_2=\sqrt{1.4\times 287\times 421.9}=\sqrt{169,519}\approx 411.7m/s$. Ratio: $a_2/a_1=411.7/316.9\approx 1.299$. Check: $\sqrt{T_2/T_1}=\sqrt{421.9/250}=\sqrt{1.6876}\approx 1.299$. ✓ Why it must match: across the shock the gas is still air ($\gamma ,R$ unchanged), so $a\propto \sqrt{T}$ holds even though the flow is violently non-isentropic. The sound-speed formula depends on the local state ($T$), not on how the gas got there. See Normal shock waves.

$R_{\mathrm{C}{\mathrm{O}}_2}=R_u/M=8314/44=188.95\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Invert $a=\sqrt{\gamma RT}$: $T=\frac{a^2}{\gamma R}=\frac{300^2}{1.289\times 188.95}=\frac{90,000}{243.55}\approx 369.5\mathrm{K}.$ For air to reach $a=300m/s$: $T_{\text{air}}=300^2/(1.4\times 287)=90,000/401.8\approx 224\mathrm{K}$. CO₂ needs a much hotter gas ($\approx 370$ K vs $224$ K). Why: CO₂ molecules are heavy (large $M$ → small $R$) and less springy (lower $\gamma$), so to make the pulse travel at the same speed you must heat them until their thermal motion matches. Heavier, floppier medium → needs more temperature for the same sound speed.

• Face of the formula to use when given $T$? ::: $a=\sqrt{\gamma RT}$
• Face to use when given $p$ and $\rho$? ::: $a=\sqrt{\gamma p/\rho }$
• $a$ scales with temperature as? ::: $a\propto \sqrt{T}$
• Same $V$, colder air → Mach number does what? ::: Increases (because $a$ drops)
• Does a normal shock invalidate $a=\sqrt{\gamma RT}$? ::: No — it is a local state property; evaluate each side separately