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

WHAT is the speed of sound?

The whole game is to compute $\partial p/\partial \rho$ for an ideal gas under the right thermodynamic constraint.

HOW we derive it — from first principles

Step 1 — Stand still on the wave (control volume trick)

A sound wave moving at speed $a$ into still gas is hard to analyze because everything is unsteady. Why switch frames? In a reference frame riding with the wave, the flow becomes steady — much easier.

In the wave frame:

• Gas enters the wave at speed $a$, with pressure $p$, density $\rho$.
• Gas leaves slightly changed: speed $a - d V$, pressure $p + dp$, density $\rho + d\rho$.

Step 2 — Conservation of mass (continuity)

Mass in = mass out across the thin wave (area $A$ cancels): $\rho\, a = (\rho + d\rho)(a - dV)$

Why this step? A wave doesn't pile up mass; whatever flows in must flow out.

Expand and drop the second-order term $d\rho\, dV$ (it's a product of two tiny things → negligible): \rho a = \rho a - \rho\, dV + a\, d\rho \;\Rightarrow\; \boxed{\rho\, dV = a\, d\rho} \tag{1}

Step 3 — Conservation of momentum

Net pressure force on the slab = rate of change of momentum (momentum flux). For the control volume: $pA - (p+dp)A = \dot m\,[(a - dV) - a]$

Why this step? Newton's 2nd law for a steady flow: force = mass flow rate $\times$ change in velocity. The mass flow rate $\dot m = \rho a A$.

-dp\,A = \rho a A\,(-dV) \;\Rightarrow\; \boxed{dp = \rho a\, dV} \tag{2}

Step 4 — Combine (1) and (2)

From (1): $dV = a\, d\rho/\rho$. Substitute into (2): $dp = \rho a \cdot \frac{a\, d\rho}{\rho} = a^2\, d\rho$ \boxed{a^2 = \frac{dp}{d\rho}} \tag{3}

Why this matters: purely from mass + momentum, the speed of sound is the square root of how much pressure changes per unit density change. This is the "stiffness/inertia" idea made exact. We have not yet said anything about temperature or $\gamma$.

Step 5 — Which process? Adiabatic, not isothermal (the crucial insight)

For an isentropic ideal gas: $p = C\,\rho^{\gamma} \quad (\text{constant } C)$

Why this relation? Combining $pV^\gamma=$ const with $V \propto 1/\rho$ gives $p \propto \rho^\gamma$. Here $\gamma = c_p/c_v$.

Differentiate: $\frac{dp}{d\rho} = C\gamma\,\rho^{\gamma-1} = \gamma\,\frac{C\rho^{\gamma}}{\rho} = \gamma\,\frac{p}{\rho}$

So a^2 = \frac{dp}{d\rho} = \gamma\,\frac{p}{\rho} \tag{4}

Step 6 — Bring in the ideal gas law

Ideal gas: $p = \rho R T$ (with $R$ the specific gas constant, $R = R_u/M$). Hence $p/\rho = RT$: $a^2 = \gamma R T \;\Rightarrow\; \boxed{\,a = \sqrt{\gamma R T}\,}$

WHY $a$ depends only on $T$ (Feynman-check)

$p$ and $\rho$ both appear, but their ratio $p/\rho = RT$ is fixed by temperature. Squeeze the gas (raise $p$ and $\rho$): the stiffness goes up but so does inertia, and they cancel. Only the random thermal molecular speed — set by $T$ — survives. In fact $a \sim$ the average molecular speed, which makes sense: sound can't outrun the molecules carrying it.

Worked examples

Common mistakes

Quick recall

Flashcards

Connections

• Mach number and flow regimes
• Isentropic relations p ∝ ρ^γ
• Ideal gas law and specific gas constant
• Compressibility and why M > 0.3 matters
• Adiabatic vs isothermal processes
• Stagnation properties and energy equation
• Normal shock waves

Concept Map

Hinglish (regional understanding)