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Calculate q,w, `DeltaU` and `DeltaH` for the reversible isothermal expansion of one mole of an ideal gas at `127^(@)C` from a volume of `10dm^(3)` to `20dm^(3)`. |
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Answer» Since, the process is isothermal, `DeltaU=DeltaH=0` From first law of thermodynamics, `DeltaU=q+w=0` `q=-w` `w=-2.30nRTlog((V_(2))/(V_(1)))` `=-2.303xx1xx8.314xx400log((20)/(10))` `=-2.303xx1xx8.314xx400xx0.3010` `=-2305.3J` (Work is one by the system) `q=-w=2305.3J` (Heat is absorbed by the system). |
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