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One mole of non-ideal gas undergoes a change of state $(2\: atm, 3.01\: L,\: 95K) \: to\: (4\: atm,\: 5L,\: 245K)$ with change in internal energy, $ \Delta U = 30\: L-atm$. The change in enthalpy of the process in $ L-atm$ is

$\begin {array} {1 1} (A)\;40 & \quad (B)\;42.3 \\ (C)\;44 & \quad (D)\;\text{Not defined because pressure is not given} \end {array}$

1 Answer

$H_2 – H_1 = U_2 – U_1 + P_2V_2 – P_1V_1 = 30 + 4 \times 5 - 2 \times 3 = 44\: L-atm.$
Ans : (C)
answered Mar 17, 2014 by thanvigandhi_1
 

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