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