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, Δ U = 30.0 L atm. The change in enthalpy ( Δ H ) of the process in L atm is

Δ H = Δ U + Δ ( P V ) = Δ U + ( P 2 V 2 - P 1 V 1 ) = 30 + ( 4 × 5 - 2 × 3 ) = 30 + 14 = 44 Latm

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ChemistryThermochemistryEnthalpy of ReactionEasy2 minPYQ_2020

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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) $\to$ (4.0 atm, 5.0 L, 245 K) with a change in internal energy, $Δ U$ = 30.0 L atm. The change in enthalpy $(Δ H)$ of the process in L atm is

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Answer:

Solution:

$Δ H = Δ U + Δ (P V)$
$= Δ U + (P_{2} V_{2} - P_{1} V_{1})$
$= 30 + (4 \times 5 - 2 \times 3)$
$= 30 + 14 = 44$ Latm

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2020

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