Consider water at 1400 kpa and 200 c. what is the specific volume (in m3 /kg)?

Vw = 2.80907 E-3 m³/Kg

Explanation:

specific volume is the inverse of density:

⇒ Vw = 1 / ρ

∴ water at:

⇒ P = 1400 KPa * ( 1000 Pa / KPa ) = 1400000 Pa

⇒ T = 200°C = 473 K

Water at these conditions is found as saturated steam, specific volume would be:

⇒ Vw = R*T / P

∴ R = 8.3144 Pa.m³/ Kg.K

⇒ Vw = (( 8.3144 ) * ( 473)) / 1400000

⇒ Vw = 2.80907 E-3 m³/Kg

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"