`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 .

Options

A. `n_(1)T_(1)+n_(2)T_(2)`
B. `(n_(1)T_(1)+n_(2)T_(2))/(T_(1)+T_(2))`
C. `(n_(2)T_(1)+n_(1)T_(2))/(n_(1)+n_(2))`
D. `(T_(1)xxT_(2))/(n_(1)xxn_(2))`

Explanation

Let final temperture be `T` then
`KE_(I) +KE_(II) =KE_("mixture")`
`(3)/(2)n_(1) RT_(1) + (3)/(2)n_(2) RT_(2) = (3)/(2) (n_(1)+n_(n_(2))RT)`
`T =(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))` .