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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 ? |
Answer» 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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