The system consists of 1.8 g of H2O in a diathermic cylinder sealed from the out

Question

The system consists of 1.8 g of H2O in a diathermic cylinder sealed from the outside by a freely movable massless piston. In the initial state, the system is completely in the liquid phase, H2O (l), at 100.oC and 1.00 atm and is in equilibrium with the surroundings. In the final state, as a result of heating the cylinder, the system is completely in the vapor phase, H2O (g), at 100.oC and 1.00 atm.

Calculate q, H and U associated with the vaporization process of 1.8 g of H2O (l), at 100.oC and 1.00 atm. Make use of the tables of thermodynamic data in your book/notes for the standard enthalpy change associated with the vaporization of 1 mol of water at 100oC and 1.00 bar. You can assume that this quantity has the same magnitude at 1 atm and 1 bar.

Explanation / Answer

Given,

Mass of water=1.8 g=1.8*10-3 kg

From the steam table,

For liquid phase at 100 and 1.00 atm

H1=419.0 kJ/kg

U1=418.9 kJ/kg

For gas phase at 100 and 1.00 atm

H2=2676 kJ/kg

U2=2506 kJ/kg

Now,

H=mass*(H2-H1)=1.8*10-3 *( 2676-419.0)= 4.062 kJ=4062 J

U= mass*(U2-U1)=1.8*10-3*(2506-418.9)=3.757 kJ=3757 J

q is the heat supplied, it means it is equal to change in enthalpy.

So, q=H=4062 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.