A) \[\Delta U=0,\,\Delta H=0\]
B) \[\Delta U=+202.6J,\,\Delta H=+202.6J\]
C) \[\Delta U=-202.6J,\,\Delta H=-202.6J\]
D) \[\Delta U=0,\,\Delta H=+202.6J\]
Correct Answer: D
Solution :
[d] \[\Delta H=\]heat given so process is isobaric. \[w=-{{P}_{ext}}(\Delta V)=-1\times 2\].liter. atm \[=-202.6J\] Hence, \[q=202.6J=\Delta H\]and \[\Delta E=q+w=0\]You need to login to perform this action.
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