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`80 K``-73 K``3 K ``20 K`
Answer : D
Solution : The rms velocity of the molecules of a gas of molecular weight M at temperature T is given by <br> `C_(ms)=sqrt((3RT)/(M))`. <br> Let `M_O" and "M_H` be molecular weights of oxygen and hydrogen `T_O" and "T_H` be the corresponding kelvin temperatures at which `C_(ms)` is same for both gases. <br> `C_(ms(o))=C_(ms(H))` <br> `sqrt((3RT_(o))/(M_(o)))=sqrt((3RT_(H))/(M_(H)))` <br> Hence, `(T_o)/(M_o)=(T_H)/(M_H)` <br> Given `T_o=273+47=320K" and "M_o=32, M_H=2" ":. T_H= (2)/(32)xx320= 20K`.
Text Solution
80 K`-73 K`3 K20 K
Answer : D
Solution : `upsilon_(rms) = sqrt((3RT_(H))/(M_H)) = sqrt((3RT_("OXY"))/(M_("OXY"))` <br> `(T_H)/(M_H) = (T_("OXY"))/(M_("OXY"))` <br> or, `T_(H) = T_("OXY") (M_H)/(M_("OXY")) = (47+273) 2/32 = 320/16` <br> `= 20K`.
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At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at 47∘ CA. 80 KB. 73 KC. 3 KD. 20 K
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