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JEE CHEMISTRY MCQ: THERMODYNAMICS
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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, ∆U=30.0 L atm. The change in enthalpy (∆H) of the process in L atm is:
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Correct option: C
44
Option C (44) is correct because it is the change in enthalpy calculated using the formula ∆H = ∆U + (P₂V₂ - P₁V₁).
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