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The magnitude of enthalpy changes for irreversible adiabatic expansion of a gas from 1L to 2L is $Δ H_{1}$ and for reversible adiabatic expansion for the same expansion is $Δ H_{2}$ Then :

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$Δ H_{1} = Δ H_{2}$ enthalpy being a state function

$Δ H_{1} > Δ H_{2}$

$Δ H_{1} < Δ H_{2}$

$Δ H_{1} = Δ E_{1}$ , and $Δ H_{2} = Δ E_{2}$ where $Δ E_{1}$ and $Δ E_{2}$ are magnitudes of change in internal energy of gas in there expansions respectively.

answer is B.

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Detailed Solution

Since, initial positions are different so, $Δ H$ will be different

Final temp in irrev. is more than rev. as $w_{irrev}$ is less in magnitude.

$\Rightarrow Δ H_{rev} > Δ H_{irrev}$

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