Question

In passing through a turbine electrical generator, 1.0 L of steam at 300°C and 3500kPa pressure expands to 55L and has its temperature lowered to 110°C. What is the pressure of the steam leaving the turbine.

Answer 1

nRT=PV (Mendeleev-Klapeyrones"s law : ideal gas equation), n - number of moles, R - gase constant 8.314 J/K*mole = 0.0821 l *atm/mol * K , T - absolutly temperature K,

P - pressure, - V - volume

nRT=PV : n-const;R-const; => P*V / T= const

P₁*V₁/T₁=P₂*V₂/T₂

3500kPa*1.0L/573,15Ċ=P² *55 L/383,15ċ

P₂=3500*383,15/(55*573,15) kPa

P=41,81kPa