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Question

Let, the average molar mass of air is 24.9 g. An open vessel at 27 oC is heated upto t oC until 1/3rd of the air measured at final temperature excapes out. The rms velocity of air molecules at t oC is: (Take R=8.3 JK1mol1)

A
340.12 m sec1
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B
420.35 m sec1
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C
515.25 m sec1
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D
632.45 m sec1
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Solution

The correct option is D 632.45 m sec1
Initial temperature =300 K
Let new temperature corresponding to toC is T K.
Let number of moles present at T K=n
Moles of air excaped at T K=n/3
So, number of moles at 300 K n+n3=4n3
Volume and pressure are kept constant
n1n2=T2T1
4n3×300=nT
T=400 K
Now, r.m.s velocity =3RTM=3×8.3×40024.9×103=632.45 m sec1.

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