Question

`16g` of oxygen at `37^(@)C` is mixed with `14g` of nitrogen at `27^(@)C`. Find the temperature of the mixture?

Answer 1

Correct Answer - `32^(@)C`
Both are diatomic.
No. of moles of `O_(2) = n_(1) = (m)/(M) = (17)/(32) = (1)/(2)`,
No. of moles of `N_(2) = n_(2) = (m)/(M) = (14)/(28) = (1)/(2)`
`:. (n_(1)+n_(2)) C_(v)T = n_(1)C_(v)T_(1) +n_(2)C_(v)T_(2)` [By energy conservation
as `C_(v)` cancel out
`=T = (n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2)) , = (((1)/(2))310+((1)/(2))300)/((1)/(2)+(1)/(2))`
`= 305 K = 32^(@)C`