Consider 100 mL of a liquid contained in an isolated container at a pressure of 1 bar. The pressure is steeply increased to 100 bars. The volume of the liquid is decreased by 1 mL at this constant pressure. Find the value of AH. (given `Delta U = 10 J)`

Options

A. `+ 680 J`
B. `+ 990 J`
C. `- 680 J`
D. `-990 J`

Explanation

We know that at constant pressure conditions
`Delta H = Delta U + Delta ( pV)`
where `Delta p V = p _(2) V_(2) - p _(1) V _(1)`
Given that `p _(1) = 1 ` bar, `V _(1) = 100 mL , p _(2) = 100 ` bar,` V _(2) = 99 mL.` substituting, we get
`Delta p V = 100 xx 99 - 1 xx 100 = 9800 mL ` bar `= 980 J (as 1 L ` bar = 100 J)
So, `Delta H = 100 + 980 = 990 J`