One mole of a non-ideal gas undergoes a change of state ( 2.0 a t o m , 3.0 L , 95 K ) → ( 4 a t m , 5 L , 245 K ) with a change in internal energy, Δ E = 30.0 L a t m . The change in enthalpy, Δ H , of the process in L atm is

Δ H = Δ E + Δ ( P V ) Δ H = 30 + 4 × 5 - 2 × 3 = 44 L a t m

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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 $(2.0 a t o m, 3.0 L, 95 K)$ $\to (4 a t m, 5 L, 245 K)$ with a change in internal energy, $Δ E = 30.0 L a t m .$ The change in enthalpy, $Δ H$ , of the process in L atm is

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Answer:

Solution:

$Δ H = Δ E + Δ (P V)$
$Δ H = 30 + (4 \times 5 - 2 \times 3) = 44 L a t m$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2020

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