The root mean square velocity of a gas molecule at 100 K and 0.5atm pressure is $106.4 {ms}^{- 1}$ If the temperature is raised to 400 K and the pressure is raised to 2atm, the root mean square velocity becomes:

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$106.4 {ms}^{- 1}$

$212.8 {ms}^{- 1}$

$425.6 {ms}^{- 1}$

$851.2 {ms}^{- 1}$

answer is C.

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Detailed Solution

It is known that;

$\begin{array}{l} U = \sqrt{\frac{3 R T}{M}} or \frac{U_{1}}{U_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \\ \frac{106.4}{U_{2}} = \sqrt{\frac{100}{400}} \end{array}$

(Note that pressure does not come in use)

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