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Two monoatomic ideal gases at temperatur...

Two monoatomic ideal gases at temperatures `T_(1)` and `T_(2)` are mixed. There is no loss of energy. If number of their moles are `n_(1)` and `n_(2)` respectively. The temperature of the mixture will be

Text Solution

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From energy conservation principle
`E_(i)=E_(f)`
`E_(1)+E_(2)=E`
`impliesn_(1)((3)/(2)RT_(1))+n_(2)((3)/(2)RT_(2))=(n_(1)+n_(2))(3)/(2)RT`
`impliesT=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`
`impliesT=(n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`
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