Please don\'t do It if you can\'t do it. Don\'t want t keep wasting questions. I
ID: 1769685 • Letter: P
Question
Please don't do It if you can't do it. Don't want t keep wasting questions. I have attached re previous solution so u can see their stupid mistakes so you don do it please don't disaaapoint me In the diagram below, the lower half of a container is occupied by water at 25° C, and the upper half is initially vacuum. The volume of the evacuated region is 3.00L. The two halves are separated by a partition. The entire containter is immersed in a thermal reservoir, also at 25 C The equilibrium vapor pressure at this temperature is 3.169 kPa (wikipedia). When the partition is removed, (a) What mass of liquid water evaporates? (Neglect the change in the volume occupied by the liquid phase.) (b) How much heat flows from the thermal reservoir into the container (in order to maintain constant temperature?) vacuum V-3.00LExplanation / Answer
on removing the partition the water evaporates to fill the void
a. volume of vacant space = 3L = 3*10^-3 m^3
Vapour pressure, P = 3169 Pa
Temperature, T = 25 C = 298.15 K
using PV = nRT
3169*3*10^-3 = n*8.31*298.15
n = 0.00383714 moles
hence n moles of water evaporates
mass of this amount of water = n*18 grams = 0.069068 grams
b. Now, latent heat of vapourisation L = (2500.8 - 2.36T + 0.0016T^2 - 0.00006T^3) J/g
for T = 25 C
L = (2500.8 - 2.36*25 + 0.0016*25^2 - 0.00006*25^3)
L = 2441.8625 J/g
hence
heat required = L*n*18 = 168.65455915 J = 169 J