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Question: The molecules of a given mass of a gas have a 'rms' velocity of \(200\;{\rm{m/s}}\) at \(27^\circ {\...

The molecules of a given mass of a gas have a 'rms' velocity of 200  m/s200\;{\rm{m/s}} at 27C27^\circ {\rm{C}} and 1×105N/m21 \times {10^5}{\rm{N/}}{{\rm{m}}^{\rm{2}}} pressure. When the temperature is 127C127^\circ {\rm{C}} and pressure is 0.5×105N/m20.5 \times {10^5}{\rm{N/}}{{\rm{m}}^{\rm{2}}}. the rms velocity in m/s{\rm{m/s}}will be

  1. 10023\dfrac{{100\sqrt 2 }}{3}
  2. 1002100\sqrt 2
  3. 4003\dfrac{{400}}{{\sqrt 3 }}
  4. 400
Explanation

Solution

Here, we will be using the formula to find the root mean square velocity. The mean square velocity is independent of the pressure. So, we use the relation between the temperature and mean square velocity.

Complete step by step answer:
Given: The initial rms velocity of molecules is v1=200  m/s{v_1} = 200\;{\rm{m/s}}, the initial temperature is T1=27C+273=300  K{T_1} = 27^\circ {\rm{C + 273 = 300}}\;{\rm{K}}, the final temperature is T2=127C+273=400  K{T_2} = 127^\circ {\rm{C + 273 = 400}}\;{\rm{K}}.
The formula for root mean square velocity is vrms=3RTM{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}
Here, R is the universal gas constant, and M is the mass of a molecule.

From, above relative, we can write vrmsT{v_{rms}} \propto \sqrt T

Now, we write the relative formula for initial and final velocity using the above relation.
v1v2=T1T2\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}}
Here, v2{v_2} is the final velocity at 127C127^\circ {\rm{C}}.
Substituting the values in above relation,

\dfrac{{200}}{{{v_2}}} = \sqrt {\dfrac{{300}}{{400}}} \\\ \dfrac{{200}}{{{v_2}}} = \sqrt {\dfrac{3}{4}} \end{array}$$ We take the square root of $4$ and get value as $2$. $$\dfrac{{200}}{{{v_2}}} = \dfrac{{\sqrt3 }}{2}$$ Now, we perform cross multiplication of $200$ and $2$ , and get $400$ as result. $$\sqrt 3 {v_2} = 400$$ Then, we divide $400$ by $\sqrt 3 $ and get the final rms velocity in the terms of meter per second. $${v_2} = \dfrac{{400}}{{\sqrt3 }}\;{\rm{m/s}}$$ Therefore, the rms velocity at $127^\circ {\rm{C}}$ is $$\dfrac{{400}}{{\sqrt3 }}\;{\rm{m/s}}$$ **So, the correct answer is “Option 3”.** **Note:** The rms velocity is the square root of the mean square of velocities of individual gas molecules. The relation between the root mean square velocity and pressure of the gas can be obtained but the root mean square velocity does not depend upon the pressure of the gas.