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First Order Reaction

A real gas is...

Question

A real gas is subjected to an adiabatic process from (5 bar, 90 L, 300 K) to (8 bar, 50 L, 300K) against a constant pressure of 6 bar. The enthalpy change for the process is

120 L-atm

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190 L-atm

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240 L-atm

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150 L-atm

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Solution

The correct option is A 190 L-atm
$q = 0, Δ U = w = - P e x t$
$(V_{2} - V_{1}) = - 6 (50 - 90) = 240 L - b a r$
$Δ H = Δ U + Δ (P V) = 240 + (8 \times 50 - 5 \times 90) = 190 L - a t m$

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