There is `1 mol` liquid (molar volume `100 ml`) in an adiabatic container initial, pressure being `1` bar Now the pressure is steeply increased to `100` bar, and the volume decreased by `1 ml` under constant pressure of `100` bar. Calculate `Delta H` and `Delta E`. [Given `1 "bar"=10^(5)N//m^(2)`]

Explanation

As the process is carriedout under adiabatic condition, `q=0` . But `DeltaU = q + w`. Hence `Delta U =w`.Further, at constant pressure of100 bar, volume has decreased by 1 ml, therefore, work of contraction .
`= P Delta V = 100 ` bar ` xx 1 `ml.
`= ( 100 xx10^(5) Nm^(-2)) ( 10^(-6)m^(3)) = 10J`
Hence, `Delta U = 10 J`
`Delta H =Delta U + Delta( PV)`
`Delta ( PV) = (P_92)V_(2) - P_(1)V_(1))`
But `P_(1) =1 "bar", V+(1) = 100 ml`
`P_(2) = 100 "bar", V_(2) = 99 ml`
`:. Delta ( PV) = ( 100 "bar" xx 99 ml) - ( 1 "bar" xx 100 ml)
`= ( 9900 - 100 ) "bar" ml = 9800 "bar ml ) = 980 J `
`:. Delta H = 10J + 980 J = 990 J`