In the given enthalpy (H) vs pressure (P) graph for reversible process ABCDA, (1 mol of ideal gas) if |2ΔSA→B+ΔSC→D|=1.4 cal/K then find ‘P2’ in barR=2 cal/K.mol and ln 2 = 0.7

DSAB=qABTAB=R17ln12TAB DSC→D=Rln2=−Rln12 2SAB+ΔSC→D=2Rln12−Rln12 =Rln12 =−Rln2 ∴P=7 bar

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In the given enthalpy (H) vs pressure (P) graph for reversible process ABCDA, (1 mol of ideal gas) if $| 2 Δ S_{A \to B} + Δ S_{C \to D} | = 1.4 c a l / K$ then find ‘ $P_{2}$ ’ in bar
$R = 2 c a l / K . m o l$ and $l n 2 = 0.7$

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Detailed Solution

$D S_{A B} = \frac{q_{A B}}{T_{A B}} = \frac{R_{17} l n \frac{1}{2}}{T_{A B}} D S_{C \to D} = R l n 2 = - R \ln \frac{1}{2} 2 S_{A B} + Δ S_{C \to D} = 2 R l n \frac{1}{2} - R l n \frac{1}{2} = R l n \frac{1}{2} = - R l n 2 ∴ P = 7 b a r$

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