Question

a volume of 1000 ml of a liquid contained in an isolated container at a pressure of 1bar. the pressure is steeply increases to 100bar by which the volume is decresed by 1ml .the change in enthalpy,of the liquid is

( answer is abcd, where a=1, if delta H= positive

and a=2 if delta H = negative)

bcd is the magnitude of delta H , in joules

ans =1990

how ?

Answer 1

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

this is the correct answer

answer in your book may be wrong so that you are confused