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Question: At what temperature pressure remaining constant will the root mean square velocity of a gas be half ...

At what temperature pressure remaining constant will the root mean square velocity of a gas be half of its velocity at 00C{0^0}C.
A. 2040C- {204^0}C
B. 68250C{6825^0}C
C. 4040C{404^0}C
D. None of the above

Explanation

Solution

Hint: The root mean square velocity is denoted by the symbol Vrms{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 00C{0^0}C .

Formula used:
VrmsT{V_{rms}} \propto \sqrt T , where Vrms{V_{rms}} is the root mean square velocity and TT is the temperature in Kelvin scale.
Tk=TC+273.15{T_k} = {T_C} + 273.15 Tk{T_k} denotes the temperature at Kelvin scale and TC{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 00C{0^0}C
We know VrmsT{V_{rms}} \propto \sqrt T , where Vrms{V_{rms}} is the root mean square velocity and TT is the temperature in Kelvin scale. At zero degrees let the root mean square velocity be Vrms{V_{rms}} . Using Tk=TC+273.15{T_k} = {T_C} + 273.15 ,we get 00C=273.15K{0^0}C = 273.15K .Therefore applying VrmsT{V_{rms}} \propto \sqrt T
Vrms273.15{V_{rms}} \propto \sqrt {273.15}
In the second case it is given that Vrms=Vrms2{V_{rms}} = \dfrac{{{V_{rms}}}}{2} and let the temperature be TT.
Vrms2T\dfrac{{{V_{rms}}}}{2} \propto \sqrt T
Taking ratio 12=T273.15\dfrac{1}{2} = \dfrac{{\sqrt T }}{{\sqrt {273.15} }}
T=273.154KT = \dfrac{{273.15}}{4}K
Using this equation Tk=TC+273.15{T_k} = {T_C} + 273.15 we get
TC=273.15(141){T_C} = 273.15\left( {\dfrac{1}{4} - 1} \right)
TC=204.860C2040C{T_C} = - {204.86^0}C \approx - {204^0}C
The correct option is A

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