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Speed of Sound (Ideal Gas)

Local speed of sound in an ideal gas as a function of temperature, used to compute Mach number.

AerodynamicsCompressible FlowFundamental

Speed of Sound (Ideal Gas) Calculator

a = √( γ · R · T )
Solve for a, γ, R, or T
a γ, R, T
m/s
J/(kg·K)
K
Solve for:
Result
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Speed of Sound vs. Temperature a(T) = √(γ · R · T)
a(T) for fixed γ, R Computed point
γ > 1, R > 0, T > 0 • a in m/s

Interpretation

Speed of sound in an ideal gas: a = √(γ·R·T), where γ is specific heat ratio, R is specific gas constant, T is absolute temperature. Example: Air (γ=1.4, R=287 J/kg·K, T=288 K) → a = √(1.4×287×288) ≈ 340 m/s.

a = sqrt(γ * R * T)
Speed of Sound (Ideal Gas)

Variables

SymbolQuantityUnit
aSpeed of soundm/s
γRatio of specific heats
RSpecific gas constantJ/(kg*K)
TStatic temperatureK

What it means

The speed of sound is the speed at which small pressure disturbances propagate through a medium. In an ideal gas, it depends only on the temperature and gas properties, not on pressure. This formula is derived from the isentropic compressibility. It is fundamental in aerodynamics, as it defines the Mach number and influences compressibility effects. The speed of sound decreases with altitude (lower temperature), which affects aircraft performance. It is also used in meteorology and acoustics. Understanding this relation is essential for any compressible flow analysis and for interpreting flight data.

Worked example

Speed of Sound – Two Examples

Real‑World
Scenario: Air at sea level T = 288 K, γ = 1.4, R = 287 J/kg·K. Find speed of sound.
ParameterValue
γ1.4
R287 J/kg·K
T288 K
1a = √(γRT) = √(1.4×287×288) = √(115,718) = 340.2 m/s
Result 340 m/s ✓ Sea level
Scenario: At 11,000 m, T = 216.5 K. Find speed of sound.
ParameterValue
T216.5 K
1a = √(1.4×287×216.5) = √(86,994) = 294.9 m/s
Result 295 m/s ✓ At altitude
Key insight: Speed of sound decreases with temperature – it's slower at higher altitudes.

Common mistakes

  • Speed of sound (ideal gas): a = √(γ·R·T).
  • γ: Specific heat ratio (dimensionless).
  • R: Specific gas constant (J/(kg·K)).
  • T: Absolute temperature (K).
  • Units: (m²/s²) → m/s.

Applications

The speed of sound in an ideal gas, a = √(γ·R·T), is a fundamental property that determines the Mach number. It varies with temperature; higher temperatures increase the speed of sound. Engineers use this formula to compute Mach numbers, to design acoustic liners, and to assess the effects of altitude on engine performance. In propulsion, the speed of sound affects the design of turbomachinery (compressor blade speeds). By understanding the speed of sound, aerospace engineers can interpret compressible flow phenomena and ensure that aircraft operate within the desired Mach regime.

  • Mach number calculation for flight test and design
  • Engine component design (compressor, turbine tip speeds)
  • Noise generation and acoustic propagation studies
  • High‑speed aerodynamic testing and instrumentation
  • Atmospheric property modelling for flight planning

Frequently Asked Questions

Q01What is the Speed of Sound (Ideal Gas) used for?
A01

It calculates the local speed of sound in an ideal gas, which is needed to compute the Mach number and to analyse compressible flows.

Q02What do the variables γ, R, and T represent?
A02

γ = specific heat ratio (dimensionless)
R = specific gas constant (J/kg·K)
T = static temperature (K)

Q03Why is the speed of sound dependent on temperature?
A03

It is proportional to the square root of temperature because sound waves propagate through molecular collisions, and higher temperature means faster molecular motion.

Q04What is the speed of sound at sea level?
A04

At 288.15 K, with γ=1.4 and R=287.05, a = √(1.4×287.05×288.15) ≈ 340.3 m/s (≈ 1116 ft/s).

Q05What are common mistakes when using this formula?
A05

  • Using sea‑level temperature at altitude instead of the actual (lower) local static temperature.
  • Using the wrong gas constant for the gas mixture (e.g., using air’s R for steam).
  • Confusing static temperature with stagnation temperature.

Q06Give a worked example.
A06

At an altitude where T = 220 K, for air γ=1.4, R=287 J/kg·K: a = √(1.4×287×220) = √(88396) ≈ 297.3 m/s.

Q07How does the speed of sound vary with altitude?
A07

It decreases with altitude in the troposphere (due to temperature decrease), then becomes roughly constant in the stratosphere (isothermal region).

Q08What is the speed of sound in other gases?
A08

For helium (γ=1.66, R=2077), at 300 K, a ≈ √(1.66×2077×300) ≈ 1017 m/s, much higher than in air.

Q09How does humidity affect the speed of sound?
A09

Humidity slightly increases the speed of sound because water vapour has a lower molar mass and higher γ? Actually, the effect is small (less than 1%).

Q10What is the relation between speed of sound and Mach number?
A10

Mach number M = V/a, so the speed of sound is the reference speed for compressibility effects.