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Given a mole of a non-ideal gas which undergoes a change of state (2.0 atm, 3.0 L 95 K) $\rightarrow$ (4.0 atm, 5.0 L, 245 K) with a change in internalenergy, $\Delta U$ = 30.0 L atm. What will be the change in enthalpy ($\Delta H$) of the process in L atm?


(a) 40.0
(b) 42.3
(c) 44.0
(d) not defined, because pressure is not constant

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Answer: 44.0
$H=U+PV$
$\Rightarrow \Delta H = \Delta U + \Delta PV$
$\Rightarrow \Delta H = \Delta U + (P_2V_2 - P_1V_1)$
$\Rightarrow \Delta H = 30+(5\times 4 -3\times 2) = 30 + (20-6) = 44$ L atm
answered Feb 28, 2014 by mosymeow_1
 

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