Let, the average molar mass of air is 24.9g. An open vessel at 27oC is heated upto toC until 1/3rd of the air measured at final temperature excapes out. The rms velocity of air molecules at toC is: (Take R=8.3JK−1mol−1)
A
340.12msec−1
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B
420.35msec−1
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C
515.25msec−1
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D
632.45msec−1
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Solution
The correct option is D632.45msec−1 Initial temperature =300K Let new temperature corresponding to toC is TK. Let number of moles present at TK=n Moles of air excaped at TK=n/3 So, number of moles at 300K⇒n+n3=4n3 Volume and pressure are kept constant ∴n1n2=T2T1 4n3×300=nT ∴T=400K Now, r.m.s velocity =√3RTM=√3×8.3×40024.9×10−3=632.45msec−1.