Let the final temperature of the mixture be T.
Since, there is no loss in energy.
ΔU=0
⇒2F1n1RΔT+2F2n2RΔT=0
⇒2F1n1R(T1−T)+2F2n2R(T2−T)=0
⇒T=F1n1R+F2n2RF1n1RT1+F2n2RT2=F1n1+F2n2F1n1T1+F2n2T2
JEE Main 2021 — Physics Thermodynamics
Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2,m1 and m2,n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is:
Held on 17 Mar 2021 · Verified 6 Jul 2026.
n1+n2n1T1+n2T2
n1F1+n2F2n1F1T1+n2F2T2
F1+F2n1F1T1+n2F2T2
n1+n2n1F1T1+n2F2T2
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