Hear that, air?

First, it is important to note two things:

1. Ideal gas has no internal interactions between particles. This means that all of the energy in the particle is kinetic energy.

   1. exp(w) = $e^{w}$

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Now, we will focus on energy range. Since the target speed is anywhere from $335$ m/s to $345$ m/s, we will have energy levels from

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$\frac{1}{2}(6.646 \times 10^{-27})(335)^{2} \approx 3.729 \times 10^{-22} J$

to

$\frac{1}{2}(6.646 \times 10^{-27})(345)^{2} \approx 3.955 \times 10^{-22} J$

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Next, we can integrate the probability density function, $p(E)$ , from the lower energy level to the higher energy level.

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$\int^{3.955 \times 10^{-22}}_{3.729 \times 10^{-22}} (\frac{2}{kT\sqrt{\pi}} \times \sqrt{\frac{E}{kT}} \times e^{-\frac{E}{kT}})\,dE$

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Solving this integral, we find the probability to be $\fbox{0.001683}$

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Relevant Quantities

$k$ is the Boltzmann constant or $1.38 \times 10^{-23}$ $m^{2}$ $kg$ $s^{-2}$ $K^{-1}$

$T$ is the temperature in Kelvins . In this case, it is $303.15 ^\circ K$

Energies in J corresponding to 335 m/s and 345 m/s are:

$E_1= \frac{1}{2} m (335^2) = 3.72924 \times 10^{-22}\\ E_2= \frac{1}{2} m (345^2) = 3.95520075 \times 10^{-22}$

Using $k=1.38 \times 10^{-23}$ , $T=273.15+30$ , $m=6.646 \times 10^{-27}$ , required probability is:

$\int_{E_1}^{E_2} p(E) \, dE=0.00168489$

# from math import * p = lambda E: (2.0/((1.380649 pow(10,-23)) 303 sqrt(pi))) * sqrt(E/((1.380649 pow(10,-23)) 303)) * pow(e,(-E/((1.380649 pow(10,-23)) 303))) ll = 3.323 pow(10,-27) 335 335 ul = 3.323 pow(10,-27) 345 345 def integrate(f, l, u, dx): ans = 0.0 x=l while x<u: ans += f(x) x += dx return ans dx integrate(p,ll,ul,pow(10,-29))

I wasn't sure whether this would work. I tried finding the integral mathematically but couldn't get anywhere. So I thought I'd try it some other day and just wrote the above out of curiosity to see if it'd work. It's not very accurate, but it works for upto 5 decimal places.

The probability can be determined by evaluating the integral of the probability density function $p(E)$ with lower and upper limits $E_1$ and $E_2$ , respectively.

$E_1 = \frac12mv_1^2 = \frac12(6.646 \times 10^{-27} kg)(335 m/s)^2 = 3.7292 \times 10^{-22} J.$ $E_2 = \frac12mv_2^2 = \frac12(6.646 \times 10^{-27} kg)(345 m/s)^2 = 3.9552 \times 10^{-22} J.$ Evaluating the integral, we get $\int\limits_{E_1}^{E_2} p(E) dE = 0.001683.$

$E = \frac{1}{2} m v^2$

Minimum energy is when $v = 335 m/s$ : $E = 3.72923675*10^-22$

Maximum energy is when $v = 345 m/s$ : $E = 3.95520075*10^-22$

Integrate probability distribution function from minimum energy to maximum energy. I did it numerically.

The lowest possible energy that the atom can have to be in our acceptable range happens with the lowest possible speed and equals $E_{\text{min}} = \frac{1}{2} mv_{\text{min}}^2 = 3.729 \cdot 10^{-22} \text{ J}$

Similarly the maximum happens with the maximum speed $E_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = 3.955 \cdot 10^{-22} \text{ J}$

Now that we have the energy bounds, we must simply integrate the PDF function in the range to find the probability that the energy (and by extension the speed) is in the desired range: $p = \int_{E_{\text{min}}}^{E_{\text{max}}} \frac{2}{kT\sqrt{\pi}} \sqrt{\frac{E}{kT}} \exp\left( -\frac{E}{kt} \right) \, dE$ where $k=1.38\cdot 10^{-23}$ J $\cdot$ K $^{-1}$ and $T = 273+30$ K.

Evaluating gives us that $p \approx \boxed{0.0001683}$ .

According to kinetic gas theory and since we are dealing with an ideal gas, the total energy for an atom equals its kinetic energy. Therefore $E=(mv^{2})/2$ , where $m$ is the mass of a helium atom. The probability that an atom has a speed near the speed of sound, which means a speed between $v_1=335m/s$ and $v_2=345m/s$ is the probability that the atom's energy is between $E_1=(mv_1^{2})/2=3.729\times10^{-22}J$ and $E_2=(mv_2^{2})/2=3.955\times10^{-22}J$ . Given that the probability as a function of energy for an atom in an ideal gas is $p(E)$ (where $p(E)$ is the function given in details and assumptions) the desired probability is the integral of $p(E)$ from $E_1$ to $E_2$ . By doing so we find that the probability is $0.00168$ , keeping 5 decimal digits. And now just a few notes:
The temperature we substitute in the $p(E)$ formula needs to be in S.I. units so $T=30+273.15=303.15K$ . That $k$ in $p(E)$ formula is Boltzmann's constant and it is equal to $1.3806488\times10^{-23}m^{2}kg(s)^{-2}K^{-1}$ .

The energy of an atom with speed 335 is $a=\dfrac{1}{2}6.646\cdot10^{-27}\cdot 335^2$ The energy of an atom with speed 345 is $b=\dfrac{1}{2}6.646\cdot10^{-27}\cdot 345^2$ If we integrate $p(E)$ from $a$ to $b$ we get our answer, $\boxed{0.001683}$ .

To find the probability that that the particle has an energy in a certain range, we simply integrate the probability density function over that range.The probability density function is a function of energy, so we must find the kinetic energies associated with the given velocities $\ v_1=335 \ \mathrm{m/s}, \ v_2=345 \ \mathrm{m/s}$ .

$$K 1 = \frac{1}{2}m v 1^2=3.746 \times 10^{-22} \ \mathrm{J}$$

$$K 2 = \frac{1}{2}m v 2^2=3.973 \times 10^{-22} \ \mathrm{J}$$

So then the probability $P$ is

$$P=\int {K 1}^{K_2} {\frac{2}{kT\sqrt{\pi}}\sqrt{\frac{E}{kT}}\exp{\left(-\frac{E}{kT}\right)}} \ dE$$

which (taking $k=1.38 \times 10^{-23} \ \mathrm{J/K}$ and $T=303 \ \mathrm{K}$ ) can be evaluated numerically to be

$$ \fbox{0.001697}. $$

The probability for an atom to have a speed in a given range is $P(v_1,v_2)=\int_{v_1}^{v_2}p(v)\,\mathrm{d}v$ . As the speed can be bijectively mapped to the kinetic energy, we can also calculate the sought probability as $P(v_1,v_2)=P(E_1,E_2)=\int_{E_1}^{E_2}p(E)\,\mathrm{d}E$ .

On the other hand, energies are expressed through speeds as $E=\frac{1}{2}mv^2$ . Thus $P(E_1,E_2)=\frac{2}{kT\sqrt{\pi}}\sqrt{\frac{E}{kT}}\int_{E_1}^{E_2}{\exp(-E/kT)}\,\mathrm{d}E$ , which, substituting the numbers, gives us $P(335,345)=0.168\%$ .