`n_(1)` and `n_(2)` moles of two ideal gases (mo1 wt `m_(1)` and `m_(2))` respectively at temperature `T_(1)K` and `T_(2)K` are mixed Assuming that no loss of energy the temperature of mixture becomes .

`n_(1)` and `n_(2)` moles of two ideal gases (mo1 wt `m_(1)` and `m_(2

1.	`n_(1)` and `n_(2)` moles of two ideal gases (mo1 wt `m_(1)` and `m_(2))` respectively at temperature `T_(1)K` and `T_(2)K` are mixed Assuming that no loss of energy the temperature of mixture becomes .
Answer» Total energy of molecules of first gas =`3/2 n_(1)kT_(1)`, Total energy of molecules of second gas =`3/2 n_(2)kT_(2)` Let temperature of mixture be `T` then total energy of molecules of mixture =`3/2 k(n_(1) + n_(2))T` `therefore (3)/(2) (n_(1)+n_(2))kT =3/2 k(n_(1)T_(1) +n_(2)T_(2)) implies T=(n_(1)T_(1) + n_(2)T_(2))/((n_(1)+n_(2)))`

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`n_(1)` and `n_(2)` moles of two ideal gases (mo1 wt `m_(1)` and `m_(2))` respectively at temperature `T_(1)K` and `T_(2)K` are mixed Assuming that no loss of energy the temperature of mixture becomes .

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