Question

107 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boiling point of water, 373.15 K. We then add heat to boil the water, converting it all to vapor.

The molecular weight of water is 18 g/mol and the latent heat of vaporization is 2260 J/g.

1) How much heat is required to boil the water?

2) Assume that the liquid water takes up approximately zero volume, and the water vapor takes up some final volume Vf. You may also assume that the vapor is an ideal gas. How much work did the vapor do pushing on the piston?

3) How much did the water internal energy change?

Answer 1

Answer:

Explanation:

1.)

$Q=mL\\\\=107g \times 2260J/g\\\\=241.82KJ$

2.)

$Moles=\frac{m}{M}=\frac{107g}{18g}=5.944moles$

using $PV=nRT\\\\V_f=\frac{nRT}{P}\\\\V_f=\frac{5.944\times 8.314\times 373.15}{101325}\\\\=0.182m^3$

At constant pressure

$W=Pa(\bigtriangleup V)\\\\Pa(V_f-0)=PaV_f\\\\=101325\times 0.182=18.44KJ$

3.)

$Q=\bigtriangleup U + W\\\\\bigtriangleup U=Q-W=241.82-18.44=223.38KJ$