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

Answer

B

Explanation

We know that the constant pressure condition,s `Delta H = DeltaU + (pV)`
where `Deltap V = p _(2) V _(2) - p _(1) V _(1)`
Given that `p _(1) = ` bar, `V _(1) = 100 mL p _(2) = 100` bar , `V _(2) = 99 mL.` Substituting, we get
`Deltap V= 100 xx 99 - 1 xx 100 = 9800 mL ` bar `= 980 J` (as 1 L bar = 100 J)
So,` Delta H = 10 + 980 = 990 J`