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Question: At what temperature is the rms speed of \({H_2}\) molecules the same as that of oxygen molecules at ...

At what temperature is the rms speed of H2{H_2} molecules the same as that of oxygen molecules at 1327C1327^\circ C?
A.173K173K
B.100K100K
C.400K400K
D.523K523K

Explanation

Solution

The speed and temperature have a unique relation in them. It varies as our condition varies. For gaseous molecules also there is a relation between the temperature and their speed. The speed of molecules in a gas is proportional to temperature and inversely proportional to molar mass.

Complete step by step answer:
As we all know that the speed of molecules in a gas is proportional to temperature and inversely proportional to molar mass.
And we also know that
Vrms=3RTM{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} where,
Vrms{V_{rms}} is root mean square velocity.
RR is the gas constant.
TT is the temperature.
MM is the molecular mass of a given gaseous species.
So using this formula we will find our answer.
We have given that Vrms{V_{rms}} of both the species, hydrogen and oxygen is equal.
Therefore,
(Vrms)O=(Vrms)H{({V_{rms}})_O} = {({V_{rms}})_H}
3RTOMO=3RTHMH\sqrt {\dfrac{{3R{T_O}}}{{{M_O}}}} = \sqrt {\dfrac{{3R{T_H}}}{{{M_H}}}} -------------------(1)
Where,
MO=32={M_O} = 32 = Molecular weight of oxygen
MH=2={M_H} = 2 = Molecular weight of hydrogen
TH=?={T_H} = ? = Temperature of hydrogen gas
TO=1327+273=1600K={T_O} = 1327 + 273 = 1600K = Temperature of oxygen (\therefore Temperature should be inKK. )
Put all these values in equation (1)
3R×160032=3RTH2\sqrt {\dfrac{{3R \times 1600}}{{32}}} = \sqrt {\dfrac{{3R{T_H}}}{2}}
TH=100K\Rightarrow {T_H} = 100K
Hence option (B) is correct, 100K100K.

Note:
Approximately in every scientific or analytical calculation temperature is taken in Kelvin (K)(K), but to confuse you the examiner can give temperature in degree Celsius (C^\circ C ) or degree Fahrenheit (F^\circ F) , so you have to convert that in Kelvin and then solve the numerical.