At what temperature will He atoms have the same $u_{ ext {rms }}$ value as $\mathrm{N}_{2}$ molecules at $25^{\circ} \mathrm{C}$ ?

Question

At what temperature will He atoms have the same $u_{	ext {rms }}$ value as $\mathrm{N}_{2}$ molecules at $25^{\circ} \mathrm{C}$ ?

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the known values**: The root-mean-square speed ($u_{ ext{rms}}$) of a gas is given by the equation $u_{ ext{rms}} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a gas particle. We know the $u_{ ext{rms}}$ for $\mathrm{N}_2$ at $25^\circ\mathrm{C}$ (which is $298\mathrm{K}$) and need to find the temperature at which He has the same $u_{ ext{rms}}$. ‖ ‖ 2 ⋮ **Set up the equation for $\mathrm{N}_2$**: For nitrogen at $25^\circ\mathrm{C}$, the equation is $u_{ ext{rms}}(\mathrm{N}_2) = \sqrt{\frac{3kT_{\mathrm{N}_2}}{m_{\mathrm{N}_2}}}$. ‖ ‖ 3 ⋮ **Set up the equation for He**: For helium, the equation is $u_{ ext{rms}}(\mathrm{He}) = \sqrt{\frac{3kT_{\mathrm{He}}}{m_{\mathrm{He}}}}$. Since we want the $u_{ ext{rms}}$ to be the same for both gases, we set the equations equal to each other: $\sqrt{\frac{3kT_{\mathrm{N}_2}}{m_{\mathrm{N}_2}}} = \sqrt{\frac{3kT_{\mathrm{He}}}{m_{\mathrm{He}}}}$. ‖ ‖ 4 ⋮ **Solve for $T_{\mathrm{He}}$**: By squaring both sides and rearranging the equation, we get $T_{\mathrm{He}} = T_{\mathrm{N}_2} imes \frac{m_{\mathrm{N}_2}}{m_{\mathrm{He}}}$. The molar mass of $\mathrm{N}_2$ is approximately $28\mathrm{g/mol}$ and for He it is approximately $4\mathrm{g/mol}$. Since the mass of a gas particle is proportional to its molar mass, we can use these values to find the temperature for He. ‖ ‖ 5 ⋮ **Calculate the temperature for He**: Plugging in the values, we get $T_{\mathrm{He}} = 298\mathrm{K} imes \frac{28}{4} = 298\mathrm{K} imes 7 = 2086\mathrm{K}$. ‖ ∻Answer∻ ⚹ The temperature at which He atoms will have the same $u_{ ext{rms}}$ value as $\mathrm{N}_2$ molecules at $25^\circ\mathrm{C}$ is $2086\mathrm{K}$. ⚹ ∻Key Concept∻ ⚹ The root-mean-square speed of gas particles is directly proportional to the square root of the temperature and inversely proportional to the square root of the particle's mass. ⚹ ∻Explanation∻ ⚹ Since the $u_{ ext{rms}}$ is the same for both gases, the temperature for He must be adjusted for the difference in molar mass between $\mathrm{N}_2$ and He.◊What is the relationship between temperature and root-mean-square velocity in gases according to the kinetic theory of gases?⍭ Generate me a similar question◊⚹

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