An open bulb containing air at 290C was cooled to a certain temperature at which the number of moles of the gaseous molecules increased by 50%. The final temperature is

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39.6 K

233.6 K

201.3 K

240 K

answer is C.

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Detailed Solution

n1T1=n2T2 n1=(xm) T1=302K n2=xM+x2M=3x2M T2=? xM×0.302=3x2M×T2⇒6003=201.3K

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