1 \(3.52 \times {10^3}\,\,K\)
2 \(2.52 \times {10^2}\,\,K\)
3 \(2.52 \times {10^4}\,\,K\)
4 \(4.52 \times {10^5}\,\,K\)
Explanation:
Root mean square speed of argon atom
\({\left( {{v_{rms}}} \right)_{Ar}} = \sqrt {\frac{{3R{T_{Ar}}}}{{{M_{Ar}}}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Root mean square speed of helium atom
\({\left( {{v_{rms}}} \right)_{He}} = \sqrt {\frac{{3R{T_{He}}}}{{{M_{He}}}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
Dividing Eq. (1) by Eq. (2)
\(\dfrac{\left(v_{r m s}\right)_{A r}}{\left(v_{r m s}\right)_{H e}}=\sqrt{\dfrac{3 R T_{A r}}{M_{A r}} \times \dfrac{M_{H e}}{3 R T_{H e}}}\)
\(\begin{gathered}\text { Given }\left(v_{r m s}\right)_{A r}=\left(v_{r m s}\right)_{H e} \\\Rightarrow 1=\sqrt{\left(\dfrac{T_{A r}}{T_{H e}}\right)\left(\dfrac{M_{H e}}{M_{A r}}\right)} \\\Rightarrow \dfrac{M_{H e}}{M_{A r}}=\dfrac{T_{H e}}{T_{A r}} \\T_{A r}=253.15\left(\dfrac{39.9}{4}\right)=2.52 \times 10^{2} \mathrm{~K} .\end{gathered}\)