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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K)With a change in internal energy ∆E=30 L atm. The change in enthalpy ∆H in the process in L-atm is

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answer is 3.

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Detailed Solution

∆H=∆E+∆PV=∆E+P2V2-P1V1 or ∆H=30+20-6 = 30+14=44 L-atm.

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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K)With a change in internal energy ∆E=30 L atm. The change in enthalpy ∆H in the process in L-atm is