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**Hint:**We first use the formula of \[{v_{rms}}\] and put the value of temperatures in it. After that we will have two equations one for 120K temperature and another for 480K.

Then we will take the ratio of both the \[{v_{rms}}\] to get a equation of both the \[{v_{rms}}\] i.e. \[{v_{rms}}\] for 120K and \[{v_{rms}}\] for 480K.

After solving the equation, we get the solution.

**Complete step by step solution**

We have to find the root mean square velocity of the gas, for this we use the formula \[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]

Where \[{v_{rms}}\] is the root-mean- square velocity, R=8.3144598 \[\;Jmo{l^{ - 1}}\] , T is the temperature, M is the mass of one mole.

Now let the root mean square velocity of the gas molecule at120K= V and

Let the root mean square velocity of the gas molecule at 480K \[ = {v_2}\]

Then \[{v_{rms}}\] at 120K i.e. V \[ = \sqrt {\dfrac{{3R{T_1}}}{M}} \] and \[{v_{rms}}\] at 480K \[ = {v_2} = \sqrt {\dfrac{{3R{T_1}}}{M}} \] . Where \[{T_1}\] =120K and \[{T_2}\] =480K . Now the ratio of the root mean square velocity at 120K to the ratio of the root mean square velocity at 480K = \[\dfrac{v}{{{v_2}}}\] .

After substituting we get \[\dfrac{v}{{{v_2}}}\] i.e.

\[\dfrac{v}{{{v_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}} \] now we put the value of \[{T_1}\] =120K and \[{T_2}\] =480K in the equation.

\[\dfrac{v}{{{v_2}}} = \sqrt {\dfrac{{120}}{{480}}} = \sqrt {\dfrac{1}{4}} \] . After solving we get \[\dfrac{v}{{{v_2}}} = \dfrac{1}{2}\]

So \[2v = {v_2}\] , therefore the root mean square speed of gas molecules 480K, will be 2v.

**So, the correct option is B.**

**Note:**Remember the formula of \[{v_{rms}}\] and also that \[{v_{rms}} \propto \sqrt T \] i.e. \[{v_{rms}}\] is dependent on temperature. If the temperature increase \[{v_{rms}}\] and vice-versa

Note that RMS velocity is used to predict how fast molecules are moving at a given temperature.

Also remember that the \[{v_{rms}}\] of a gas module is generated from the thermal energy equation i.e. \[KE = \dfrac{3}{2}RT\] .

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