At what temperature root mean square speed of ozone will be equal to root mean square speed of oxygen at 27c?
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Answer:
Root mean square speed of Argan atom
( Vrms1 ) = √{3RT1/M1} -----(1)
Root mean square spped of Helium atom ( Vrms2) = √ { 3RT2/M2}-------(2)
Divide equations (1) and (2)
Vrms1/Vrms2 = √{3RT1/M1}/√{3RT2/M2}
= √{ T1 ×M2/T2×M2}
But ,
Vrms1 = Vrms2 [ A/C to question , ]
1 = √{ T1 × M2/M1 × T2}
M1/M2 = T1/T2
So,
T1 = T2 × { M1/M2}
Here,
T2 = -20°C = -20+273 = 253 K
M1 = 40g/mol
M2 = 4 g/mol
T1 = ?
T1 = 253 × { 40 /4 }
= 2530 K
Explanation:
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