106 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.

The missing part of the question is;

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?

A) Q = 239.55 KJ

B) W = 18.238 KJ

C) ΔU = 221.31 J

Explanation:

We are giving;

Mass; m = 106 g

Latent heat of vaporization; L = 2260 J/g.

Molecular weight of water; M = 18 g/mol

Pressure; P = 101325 Pa

Temperature; T = 373.15 K

A) Formula for amount of heat required is;

Q = mL

Q = 106 x 2260

Q = 239560J = 239.55 KJ

B) number of moles; n = m/M

n = 106/18

n = 5.889 moles

Now, we know from ideal gas equation that;

PV = nRT

Thus making V the subject, we have; V = nRT/P

However, here we are told V is V_f. Thus, V_f = nRT/P

R is gas constant = 8.314 J/mol·K

Plugging in the relevant values ;

V_f = (5.889 x 8.314 x 373.15)/101325

V_f = 0.18 m³

Now, at constant pressure, work done is;

W = P(ΔV)

W = P(V_f - V_i)

W = 101325(0.18 - 0)

W = 101325 x 0.18

W = 18238.5J = 18.238 KJ

C) Change in water internal energy is gotten from;

ΔU = Q - W

Thus, ΔU = 239.55 KJ - 18.238 KJ

ΔU = 221.31 J