Question

The properties of a closed system change following the relation between pressure and volume as p V = 3.0 where p is in bar V is in m3. Calculate the work done when the pressure increases from 1.5 bar to 7.5 bar.

Answer 1

W=∫_v1^v2 pdV

V₁ = 3.0/p₁ = 3.0/1.5 = 2 m³

V₂ = 3.0/p₂ = 3.0/7.5 = 0.4 m³

W=∫₂^0.4 (3.0/V) dV=3×[lnV]₂^0.4=3×(ln0.4-ln2)= -4.83 barr×m³

As 1 barr = 10⁵ N/m², then W = -4.83*10⁵ (N/m²)*m³ = -4.83*10⁵ Nm = -4.83*10⁵ J = -483 kJ

Answer: -483 kJ