Question
The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at $$300\,K$$ is :
Atomic masses : $$He = 4\,u,\,O = 16\,u$$ )
A.
300$$\,K$$
B.
600$$\,K$$
C.
1200$$\,K$$
D.
2400$$\,K$$
Answer :
2400$$\,K$$
Solution :
$$\eqalign{
& {V_{rms}} = \sqrt {\frac{{3\,RT}}{M}} \cr
& {V_{rms\left( {{O_2}} \right)}} = {V_{rms\left( {He} \right)}} \cr
& \sqrt {\frac{{3\,R{T_{{O_2}}}}}{{{M_{{O_2}}}}}} = \sqrt {\frac{{3\,R{T_{He}}}}{{{M_{He}}}}} \cr
& {\text{or}}\,\,\,\frac{{{T_{{O_2}}}}}{{{M_{{O_2}}}}} = \frac{{{T_{He}}}}{{{M_{He}}}} \cr
& \therefore \,\,{T_{{O_2}}} = \frac{{300 \times 32}}{4} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2400K \cr} $$