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?
Answer:
A) Q = 239.55 KJ
B) W = 18.238 KJ
C) ΔU = 221.31 J
Step-by-step 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