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Three perfect gases at absolute temperat...

Three perfect gases at absolute temperature `T_(1), T_(2)` and `T_(3)` are mixed. The masses f molecules are `m_(1), m_(2)` and `m_(3)` and the number of molecules are `n_(1), n_(2)` and `n_(3)` respectively. Assuming no loss of energy, the final temperature of the mixture is

A

`((T_1 + T_2 + T_3))/3`

B

`(n_1 T_1 + n_2 T_2 + n_3 T_3)/(n_1 + n_2 + n_3)`

C

`(n_1 T_1^2 + n_2 T_2^2 + n_3 T_3^2)/(n_1T_1+ n_2 T_2 + n_3 T_3)`

D

`(n_1^2 T_1^2 + n_2^2T_2^2 + n_3^2 T_^3^2)/(n_1 T_1 + n_2 T_2 + n_3T_3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `T_3gt T_2 gt T_1` and final temperature is T such that `T_3lt T lt T_2 lt T_1`. Now heat gained by first two gases is equal to heat lost by third gas :
`n_1C(T-T_1)+n_2C(T-T_2)=n_3C(T_3-T) implies T= (n_1T_1+n_2T_2+n_3T_3)/(n_1+n_2+n_3)`.
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