Question

The root mean square speed of molecules of a given mass of a gas at 27⁰C and 1 atmosphere pressure is 200 ms^-1. The root mean square speed of molecules of the gas at 127⁰C and 2 atmosphere pressure is x/√3 ms^-1. The value of x will be __________.

Answer 1

Step 1: Given data

Temperature ${\mathrm{T}}_1=27°\mathrm{C}$

Temperature ${\mathrm{T}}_2=127°\mathrm{C}$

Pressure ${\mathrm{P}}_1=1\mathrm{atm}$

Pressure ${\mathrm{P}}_2=2\mathrm{atm}$

${\mathrm{V}}_{\mathrm{rms}}=200{\mathrm{ms}}^{-1}$

Step 2: Calculating the value of x

We know that,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{RT}}_1}{\mathrm{M}}} \\ \Rightarrow 200=\sqrt{\frac{3\times \mathrm{R}\times 300}{\mathrm{M}}}..................\left(\mathrm{i}\right) \end{gathered}$$

Also, it is given that

$\frac{\mathrm{x}}{\sqrt{3}}=\sqrt{\frac{3\times \mathrm{R}\times 400}{\mathrm{M}}}..................\left(\mathrm{ii}\right)$

Dividing (i)/(ii) we get,

$$\begin{gathered} \frac{200}{\left({\displaystyle \frac{\mathrm{x}}{\sqrt{3}}}\right)}=\sqrt{\frac{300}{400}} \\ \Rightarrow \frac{200}{\left({\displaystyle \frac{\mathrm{x}}{\sqrt{3}}}\right)}=\sqrt{\frac{3}{4}} \\ \Rightarrow 200=\frac{1}{2}\sqrt{3}\times \left(\frac{\mathrm{x}}{\sqrt{3}}\right) \\ \Rightarrow \mathrm{x}=400 \end{gathered}$$

Therefore, The value of x will be $400.$