Verified Solution • 2025–26 Session moderate

QJEEM21Phe0e8: JEE Main 2021 Physics - Class 11 JEEM21Phe0e8 Solution: Two ideal polyatomic gases at temperatures T 1 and T 2 are…

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 :

Multiple Choice Options

  • A n1F1T1+n2F2T2F1+F2{{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{F_1} + {F_2}}}
  • B n1F1T1+n2F2T2n1F1+n2F2{{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1}{F_1} + {n_2}{F_2}}}
  • C n1T1+n2T2n1+n2{{{n_1}{T_1} + {n_2}{T_2}} \over {{n_1} + {n_2}}}
  • D n1F1T1+n2F2T2n1+n2{{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1} + {n_2}}}

Step-by-Step Verified Solution

Initial internal energy = Final internal energy

F12n1RT1+F22n2RT2=F12n1RT+F22n2RT{{{F_1}} \over 2}{n_1}R{T_1} + {{{F_2}} \over 2}{n_2}R{T_2} = {{{F_1}} \over 2}{n_1}RT + {{{F_2}} \over 2}{n_2}RT

⇒\Rightarrow T=F1n1T1+F2n2T2F1n1+F2n2T = {{{F_1}{n_1}{T_1} + {F_2}{n_2}{T_2}} \over {{F_1}{n_1} + {F_2}{n_2}}}

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