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Q.

Three perfect gases at absolute temperatures T1,  T2    and   T3 are mixed. The masses  of the molecules are  m1,  m2 and m3 and number of molecules are n1,  n2 and n3  respectively. Assuming no loss of energy, the final temperature of the mixture is

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a

T1+T2+T33

b

n1T1+n2T2+n3T3n1+n2+n3

c

n1T12+n2T22+n3T32n1T1+n2T2+n3T3

d

n12T12+n22T22+n32T32n1T1+n2T2+n3T3

answer is B.

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Detailed Solution

Let  T3>T2>T1 and final temperature be T such that T3>T>T2>T1 .Heat lost by first two gases is equal to heat gained by third gasn1CT−T1+n2CT−T2=n1CT3−T⇒T=n1T1+n2T2+n3T3n1+n2+n3
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