At what temperature would the root-mean square speed of a gas mol… - Vidyadip

Question

At what temperature would the root-mean square speed of a gas molecule have twice its value at 100°C?

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Answer

We know that$\text{C}^2=3\frac{\text{RT}}{\text{Nm}}=3\frac{\text{Kt}}{\text{m}}$
Thus $\text{C}^2_1=\frac{3\text{kT}_1}{\text{m}}$ and $\text{C}_2^2=\frac{3\text{kT}_2}{\text{m}}$$\therefore\frac{\text{C}^2_1}{\text{C}^2_2}=\frac{\text{T}_1}{\text{T}_2}$
Here $\text{C}_2=2\text{C}_1,\text{T}=273+100=373\text{K}$$\text{T}_2=\text{T}_1\times\frac{\text{C}^2_2}{\text{C}^2_1}$
$=373\times4=1492\text{K}=1219^{\circ}\text{C}$

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