Question

100 mL of a liquid contained in an isolated 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 the ∆H and ∆U.

Answer 1

∆=+ UqW

For adiabatic process, q = 0, hence ∆U= W

W = – p( ∆V ) = – p(V₂ – V₁)

⇒∆U= –100 (99 – 100) = 100 bar mL

∆H =∆U +∆(pV)

where, ∆_pV = p₂ V₂ – p₁ V₁

∆H = 100 + (100 × 99 – 1 × 100) = 9900 bar mL