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1 L of air weighs 1.293 g at 0^(@)C and ...

1 L of air weighs 1.293 g at `0^(@)C` and 1 atm pressure. At becomes `30 dm^(3)` . To what temperature the gas must be raised to accomplish the change ?

Text Solution

Verified by Experts

Given `V_(1)=1L`
`T_(1)=0^(@)C` or `273.15K`
`P_(1)=1atm`
`V_(2)=1L`
`T_(2)=?`
`P_(2)=1atm`
`thereforeT_(2)T_(1)//V_(1)=79.9^(@)C`
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