The temperature at which the RMS speed of atoms in neon gas is equal to the RMS speed of hydrogen molecules at

Question

The temperature at which the RMS speed of atoms in neon gas is equal to the RMS speed of hydrogen molecules at $15^\circ	ext{C}$ is: (the atomic mass of neon $= 20.2 	ext{ u}$, molecular mass of hydrogen $= 2 	ext{ u}$)

Accepted Answer

The root mean square (RMS) speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$, where $T$ is the absolute temperature and $M$ is the molar mass.

1.  **Condition:** $v_{rms}(	ext{Ne}) = v_{rms}(	ext{H}_2)$
    $\sqrt{\frac{3RT_{	ext{Ne}}}{M_{	ext{Ne}}}} = \sqrt{\frac{3RT_{	ext{H}_2}}{M_{	ext{H}_2}}}$

2.  **Simplify:** Squaring both sides and canceling constants ($3R$):
    $\frac{T_{	ext{Ne}}}{M_{	ext{Ne}}} = \frac{T_{	ext{H}_2}}{M_{	ext{H}_2}}$
    $T_{	ext{Ne}} = T_{	ext{H}_2} 	imes \frac{M_{	ext{Ne}}}{M_{	ext{H}_2}}$

3.  **Substitute Values:**
    *   $T_{	ext{H}_2} = 15^\circ	ext{C} = 15 + 273 = 288 	ext{ K}$
    *   $M_{	ext{Ne}} = 20.2 	ext{ u}$
    *   $M_{	ext{H}_2} = 2 	ext{ u}$
    $T_{	ext{Ne}} = 288 	imes \frac{20.2}{2} = 288 	imes 10.1$
    $T_{	ext{Ne}} \approx 2908.8 	ext{ K}$

4.  **Scientific Notation:**
    $2908.8 	ext{ K} \approx 2.9 	imes 10^3 	ext{ K}$

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