100 mL of a liquid is contained in an insulated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1 mL at this constant pressure. Find ∆H and ∆U.
p1 = 1 bar,p2 =100bar, V1 =100mL,V2 =99mL
For adiabatic process, q = 0, ∆ U = q + W
∆U = W
W = -p∆V = -100(99 -100) = 100 bar mL
∆H = ∆U + ∆pV
= 100 + p2V2 - p1V1 = 100+ (100x99)-(1 x 100)
= 100 + 9900 -100 = 9900 bar mL
You bloody scrowndrel 
.heat measured at constant pressure is known as enthalpy.but u are treating it as variable and u r caluculating change in pressure and volume sulmtaneously.