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)`]

Answer

`Delta H=990, Delta E=19 J`

Explanation

`Delta U= q+W`
for adiabatic process `q=0`, hence `Delta U=W` and `W= -p(Delta V)= -P(V_(2)-V_(1))`
Now `Delta H= Delta U+Delta (PV)`
Here `Delta U` already calculated abobe and
`Delta PV=(P_(2)V_(2)-P_(1)V_(1))`
So, `Delta H=100+(100xx99-1xx100)=9900 ` bar `mL=990J`