JEE Main 17 March 2021 Shift 1 Solved Paper

According to question, two ideal atomic gases at temperatures T1 and T2 are mixed.
Let the final temperature of this mixture be T.
As per question there is no loss of energy, it means
∆U=0...(i)
As, we know, ∆U=‌

f1n1R∆T

2

+‌

f2n2R∆T

2

....(ii)
From Eqs. (i) and (ii), we get
‌

f1n1R∆T

2

+‌

f2n2R∆T

2

=0
⇒‌‌f1n1R∆T+f2n2R∆T=0
⇒‌‌f1n1R(T1−T)+f2n2R(T2−T)=0
⇒‌‌f1n1(T1−T)+f2n2(T2−T)=0
⇒‌‌f1n1T1−f1n1T+f2n2T2−f2n2T=0
⇒‌‌f1n1T+f2n2T=f1n1T1+f2n2T2
⇒T(f1n1+f2n2)=f1n1T1+f2n2T2
T=‌

f1n1T1+f2n2T2

f1n1+f2n2