JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 3)
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 :
$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{F_1} + {F_2}}}$$
$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1}{F_1} + {n_2}{F_2}}}$$
$${{{n_1}{T_1} + {n_2}{T_2}} \over {{n_1} + {n_2}}}$$
$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1} + {n_2}}}$$
Explanation
Initial internal energy = Final internal energy
$${{{F_1}} \over 2}{n_1}R{T_1} + {{{F_2}} \over 2}{n_2}R{T_2} = {{{F_1}} \over 2}{n_1}RT + {{{F_2}} \over 2}{n_2}RT$$
$$ \Rightarrow $$ $$T = {{{F_1}{n_1}{T_1} + {F_2}{n_2}{T_2}} \over {{F_1}{n_1} + {F_2}{n_2}}}$$
$${{{F_1}} \over 2}{n_1}R{T_1} + {{{F_2}} \over 2}{n_2}R{T_2} = {{{F_1}} \over 2}{n_1}RT + {{{F_2}} \over 2}{n_2}RT$$
$$ \Rightarrow $$ $$T = {{{F_1}{n_1}{T_1} + {F_2}{n_2}{T_2}} \over {{F_1}{n_1} + {F_2}{n_2}}}$$
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