91Ó°ÊÓ

$T=20.0°\text{C}=293\text{‿é}$

Find the RMS speed from its formula in terms of R, temperature, and molar mass.

Formula is as follow:

$v=\sqrt{\frac{3RT}{M}}$

Here, M is mass, Tis temperature, v is velocity and R is universal gas constant.

The RMS speed of nitrogen molecules is,

$v=\sqrt{\frac{3RT}{M}}$

Molar mass of nitrogen is,$M=28×{10}^{−3}\frac{\text{kg}}{\text{mol}}$

$$\begin{gathered} ∴v=\sqrt{\frac{3\left(8.314\right)\left(293\right)}{28×{10}^{−3}}} \\ \begin{array}{c}v=510.88\\ ≈511\text{ }\frac{\text{m}}{\text{s}}\end{array} \end{gathered}$$

Hence, the RMS speed of a nitrogen molecule at 20.0is $511\text{ }\frac{\text{m}}{\text{s}}$.

Consider the formula:

$\frac{v}{v^{\text{'}2}}=\frac{T}{T\text{'}}$ ….. (i)

Substitute the values and solve as:

$$\begin{gathered} \frac{{\left(1\right)}^2}{{\left(\frac{1}{2}\right)}^2}=\frac{293}{T\text{'}} \\ \begin{array}{c}{T}^{\text{'}}=73.25\text{‿é}\\ =−199.75°\text{C}\\ ≈−200°\text{C}\end{array} \end{gathered}$$

Hence,the temperatures at which the RMS speed be half that valueis $−200°\text{C}$.

Substitute the values in equation (i) and solve as:

$$\begin{gathered} \frac{{\left(1\right)}^2}{2^2}=\frac{293}{T\text{'}} \\ \begin{array}{c}{T}^{\text{'}}=1172\text{‿é}\\ =899°\text{C}\end{array} \end{gathered}$$

Hence,the temperatures at which the RMS speed be twice that valueis $899°\text{C}$.