Maxwell-Boltzmann law:

Exercise 2

The diagram represents the probability density of the speed of helium atoms at different temperatures. Check the values for $v=750\frac{m}{s}$.

$4 \pi(\frac{M}{2\pi\cdot R\cdot T})^\frac{3}{2}v^2e^{-\frac{Mv^2}{2RT} } \,=$ $4 \cdot 3.14 (\frac{0.004}{2\cdot 3.14\cdot 8.3\cdot 150})^\frac{3}{2} 750^2 e^{-\frac{0.004\cdot 750^2}{2\cdot 8.3\cdot 150} } 10\,=$ $4 \pi(\frac{M}{2\pi\cdot R\cdot T})^\frac{3}{2}v^2e^{-\frac{Mv^2}{2RT} } \,=$ $4 \cdot 3.14 (\frac{0.004}{2\cdot 3.14\cdot 8.3\cdot 300})^\frac{3}{2} 750^2 e^{-\frac{0.004\cdot 750^2}{2\cdot 8.3\cdot 300} } 10\,=$ $4 \pi(\frac{M}{2\pi\cdot R\cdot T})^\frac{3}{2}v^2e^{-\frac{Mv^2}{2RT} } \,=$ $4 \cdot 3.14 (\frac{0.004}{2\cdot 3.14\cdot 8.3\cdot 600})^\frac{3}{2} 750^2 e^{-\frac{0.004\cdot 750^2}{2\cdot 8.3\cdot 600} } 10\,=$ $4 \pi(\frac{M}{2\pi\cdot R\cdot T})^\frac{3}{2}v^2e^{-\frac{Mv^2}{2RT} } \,=$ $4 \cdot 3.14 (\frac{0.004}{2\cdot 3.14\cdot 8.3\cdot 800})^\frac{3}{2} 750^2 e^{-\frac{0.004\cdot 750^2}{2\cdot 8.3\cdot 800} } 10\,=$ $4 \pi(\frac{M}{2\pi\cdot R\cdot T})^\frac{3}{2}v^2e^{-\frac{Mv^2}{2RT} } \,=$ $4 \cdot 3.14 (\frac{0.004}{2\cdot 3.14\cdot 8.3\cdot 1000})^\frac{3}{2} 750^2 e^{-\frac{0.004\cdot 750^2}{2\cdot 8.3\cdot 1000} } 10\,=$