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One mole of non-ideal gas undergoes a change of state (`2.0 "atm"`, `3.0 L`, `95 K`) `to` (`4.0 "atm"`, `5.0 L`, `245 K`) with a change in internal energy, `DeltaU=30.0 L atm`. The change in enthalpy `(DeltaH)` of the process in `L "atm"` isA. `40.0`B. `42.3`C. `44.0`D. not defined because pressure is not constant |
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Answer» Correct Answer - C `H=U+PV` `DeltaH=DeltaU+DeltaPV` `DeltaH=DeltaU+(P_(2)V_(2)-P_(1)V_(1))` `=30+(5xx4-3xx2)` `=30+(20-6)=44 L` atm |
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