Answer: (a). T = 38.2 °C (b). V = 1.3392 cm³ (c). ii and iii
Explanation:
this is quite easy to solve, i will give a step by step analysis to solving this problem.
(a). from the question we have that;
the Mole fraction of Nitrogen, yи₂ = 0.1
Also the Mole fraction of Water, yн₂o = 0.1
We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.
ρн₂o = ṗн₂o
= yн₂oP = 0.1 × 500 mmHg = 50 mmHg
from using Antoine equation, we apply the equation
logρн₂o = A - B/C+T
T = B/(A - logρн₂o) - C
= 1730.63 / (8.07131 - log 50mmHg) - 223.426
T = 38.2 °C
We have that the temperature for the first drop of liquid form is 38.2 °C
(b). We have to calculate the total moles of gas mixture in a 30 litre flask;
n = PV/RT
n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol
Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol
Moles of N₂ is 0.9(0.744) = 0.6696 mol
we have that the moles of water condensed is 0.0744 mol i.e the water vapor in the flask is condensed
Vн₂o = 0.0744 × 18 / 1 (g/cm³)
Vн₂o = 1.3392 cm³
Therefore, the volume of the liquid water is 1.3392 cm³
(3). (ii) and (iii)
The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.
cheers i hope this helps!!!!
