As far as I know the temperature of the air depends on how fast the airmolecules are moving. But what is the increase of speed (in km/h) of those air molecules?
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$\begingroup$ The energy of a molecule follows the equi-partition theorem, $E = n k_B T$ where n is degrees of freedom of the molecules. This can then be placed into the formula for kinetic energy $E = \frac{1}{2} m v^2$. Solving for the difference of the velocities at two different temperatures gets you your answer. $\endgroup$– TweejMay 21, 2016 at 18:33
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$\begingroup$ $<v^2> \alpha T$, where $T$ is the absolute temperature, so you need to convert $T (^O C)$ to $T (K)$. $\endgroup$– jimMay 21, 2016 at 19:26
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$\begingroup$ Should find for air (remember air is a mixture of gases) at room temperature ($20^0 C$), $\sqrt{<v^2>} \approx 500 m/s$ $\endgroup$– jimMay 21, 2016 at 20:15
2 Answers
Looking around, the root mean square speed of air at $20$ C is about $500 m/s$, and given that you have $\langle v^2 \rangle \propto \, T$ so that $v_{rms}(T) = \sqrt{\langle v^2\rangle}$ varies with $\sqrt{T}$ then have $$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$ and $$v_{rms}(25) = v_{rms}(20)\times \frac{\sqrt{25+273}}{\sqrt{20+273}} \approx 504 m/s$$
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$\begingroup$ 1785 km/h - 1814 = 29 km/h. So a temperature raise of 10 degrees has an increase of speed of 29km/h or 1,6%. That is not a very big difference comparing to what we feel. $\endgroup$– MarijnMay 21, 2016 at 20:44
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1$\begingroup$ People are good at noticing temperature changes. $\endgroup$– jimMay 21, 2016 at 20:46
The speed of sound, which is effectively a measure of how fast particles are moving on average in a gas, is given by: $$ C_{s}^{2} = \frac{\partial P}{\partial \rho} \tag{1} $$ where $P$ is the pressure and $\rho$ is the mass density. For an ideal gas, one can use an adiabatic equation of state such that Equation 1 goes to: $$ C_{s}^{2} = \frac{ \gamma \ P }{ \rho } \tag{2} $$ where $\gamma$ is the adiabatic or polytrope index of the gas. For a monatomic ideal gas, one can use $\gamma = 5/3$ and the ideal gas law for pressure. Then Equation 2 goes to: $$ \begin{align} C_{s}^{2} & = \frac{ \gamma \ n \ k_{B} \ T }{ \rho } \tag{3a} \\ & = \frac{ \gamma \ k_{B} \ T }{ m } \tag{3b} \\ & = \frac{ \gamma \ R \ T }{ M } \tag{3c} \end{align} $$ where $n$ is the number density, $k_{B}$ is Boltzmann's constant, $R$ is the gas constant, $T$ is the average temperature, $m$ is the particle mass, and $M$ is the molar mass of the gas.
The Earth's atmosphere at sea level has a molar mass of ~28.9645 g $mol^{-1}$, $R = 8.3144598(48) \ J \ K^{-1} \ mol^{-1}$, we chose $\gamma = 5/3$ (or we could use 7/5 for a diatomic molecule, but it's just a constant), and now we choose temperatures.
The constant factors in front are roughly equal to: $$ \sqrt{\frac{ \gamma \ R }{ M }} \sim 21.873 $$ so that we can say the speed of sound goes as: $$ C_{s} \sim 21.873 \ \sqrt{T} \tag{4} $$ Then for we choose a bunch of temperatures and calculate the corresponding speeds of sound, which are shown in the table below.
Temperature [K] | Temperature [deg C] | Speed [m/s] | Speed [km/h] ------------------------------------------------------------------- 100.00 | -173.15 | 218.73 | 878.4 225.00 | -48.15 | 328.10 | 1181.2 273.15 | 0.00 | 361.50 | 1301.4 288.15 | 15.00 | 371.29 | 1336.6 298.15 | 25.00 | 377.68 | 1359.6 625.00 | 351.85 | 546.83 | 1968.6 900.00 | 626.85 | 656.19 | 2362.3 2500.0 | 2226.85 | 1093.65 | 3937.1
Note that if we had used $\gamma = 7/5$, the speeds would have changed by a factor of ~0.92 (i.e., decreased by roughly 8%).