ThermodynamicsHard
Question
Heat of combustion of ethanol at constant pressure and at temperature TK (= 298 K) is found to be - q J mol-1. Hence, heat of combustion (in J mol-1) of ethanol at the same temperature at constant volume will be:
Options
A.RT - q
B.- (q + RT)
C.q - RT
D.q + RT
Solution
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)
ᐃng = 2 - 3 = - 1
so ᐃU = ᐃH - ᐃng RT
= - q + RT
ᐃng = 2 - 3 = - 1
so ᐃU = ᐃH - ᐃng RT
= - q + RT
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