Question

At what temperature pressure remaining constant will the root mean square velocity of a gas be half of its velocity at $ {0^0}C $.
A. $- {204^0}C $
B. $ {6825^0}C $
C. $ {404^0}C $
D. None of the above

Answer 1

Hint: The root mean square velocity is denoted by the symbol $ {V_{rms}} $ .The root mean square velocity is proportional to the root of temperature in Kelvin scale. We have to use this detail to find the value of temperature at which the root mean square velocity will be half as that of which it will be in ${0^0}C $.

Formula used:
 $ {V_{rms}} \propto \sqrt T $ , where $ {V_{rms}} $ is the root mean square velocity and $ T $ is the temperature in Kelvin scale.
$ {T_k} = {T_C} + 273.15 $ $ {T_k} $ denotes the temperature at Kelvin scale and $ {T_C} $ denotes the temperature at Celsius scale.

Complete step-by-step answer:
We have use this detail to find the value of temperature at which the root mean square velocity will be half as that of which it will be in $ {0^0}C $
We know $ {V_{rms}} \propto \sqrt T $ , where $ {V_{rms}} $ is the root mean square velocity and $ T $ is the temperature in Kelvin scale. At zero degrees let the root mean square velocity be $ {V_{rms}} $ . Using $ {T_k} = {T_C} + 273.15 $ ,we get $ {0^0}C = 273.15K $ .Therefore applying $ {V_{rms}} \propto \sqrt T $
 $ {V_{rms}} \propto \sqrt {273.15} $
In the second case it is given that $ {V_{rms}} = \dfrac{{{V_{rms}}}}{2} $ and let the temperature be $T$.
 $ \dfrac{{{V_{rms}}}}{2} \propto \sqrt T $
Taking ratio $ \dfrac{1}{2} = \dfrac{{\sqrt T }}{{\sqrt {273.15} }} $
 $ T = \dfrac{{273.15}}{4}K $
Using this equation $ {T_k} = {T_C} + 273.15 $ we get
 $ {T_C} = 273.15\left( {\dfrac{1}{4} - 1} \right) $
 $ {T_C} = - {204.86^0}C \approx - {204^0}C $
The correct option is A

Note: We know $ {V_{rms}} \propto \sqrt T $. This can be also written as $ {V_{rms}} = k\sqrt T $, where $ k $ is a constant of proportionality. From the total molecular kinetic energy we found that $ k = \sqrt {\dfrac{{3R}}{M}} $ .Substituting this we get $ {V_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$, where $M$ denotes the molar mass, $ T $ is the temperature in Kelvin scale and $ R $ is the universal gas constant.