at what temperature will the root mean square velocity of molecules of gas be twice (Vrms)at 0°C,at what temp 1.5 times.

Dear Student,

Please find below the solution to the asked query:

The equation for the rms speed is,

$V_{rms}=\sqrt{\frac{3RT}{m}}$
Where,
R = Universal gas constant
T = absolute temperature
m = molar mass of the gas

So, at T = 300 K, rms speed is V₁, then let the temperature T₂, where the rms speed is 2V_​1.
Therefore,

$\frac{V_2}{V_1}=\frac{\sqrt{{\displaystyle \frac{3R{T}_2}{m}}}}{\sqrt{{\displaystyle \frac{3RT}{m}}}} \Rightarrow \frac{2V}{V}=\sqrt{\frac{T_2}{300}} \Rightarrow T_2=1200 K$
So, at T = 1200 K, the rms speed is double to the rms speed at T = 300 K.

Let the temperature T₃, where the rms speed is 1.5 times to the rms speed at T = 300 K.

So,

$$\begin{gathered} \frac{V_3}{V_1}=\sqrt{\frac{T_3}{300}} \Rightarrow \frac{1.5V}{V}=\sqrt{\frac{T_3}{300}} \Rightarrow T_3=2.25\times 300 \\ \Rightarrow T_3=675 K \end{gathered}$$

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