PI1. (Sec. 11.8) The vapor pressure of the organie liquid toluene at 20.0 °C is 22 is the new temperature, in °C, if the vapor pressure is increased 42 mmHg? (All\",-40.6 kJ/mol) & (R-8314 J / mol . K) mm H g What a) 24.2 °C b) 25.6 °C c) 28.1 °C d) 31.8 °C e) 39.5 °C P12. (Sec. 11.8) The vapor pressure of the organic solid naphthalene at 25 °C is 0.087 mm Hg. What is the new temperature, in °C, if the vapor pressure is carefully increased to 0.287 mmHg? AHvap 56.1 kJ/mol) & (R- 8.314J/mol K) a) 11.9°C b) 41.5 °C c) 23.4°C d) 35.4 °C e) 50.2 °c P13. (Sec. 11.8) The heat of vaporization (AHap)for water is 40.6 kJ/mol. How much energy, in kJ, is needed to convert 55.5 grams of liquid water to steam at 100 °C? a) 67.1 kJ b) 98.1 kJ c) 202 kJ d) 76.9 kJ e) 125 kJ P14. (Sec. 11.9) On the phase diagram shown below, segmentcorresponds to the conditions of temperature and pressure under which the solid and the gas of the substance are in equilibriumm
Solution
11)
T1 = 20.0 oC
=(20.0 + 273)K
= 293.0 K
P1 = 22 mmHg
P2 = 42 mmHg
?H = 40.6 KJ/mol
= 40600.0 J/mol
use:
ln(P2/P1) = (?H/R)*(1/T1 - 1/T2)
ln(42.0/22) = (40600.0/8.314)*(1/293.0 - 1/T2)
0.6466 = 4883.3293*(1/293.0 - 1/T2)
T2 = 305 K
= (305-273) oC
= 32 oC
Answer: d
12)
T1 = 25.0 oC
=(25.0 + 273)K
= 298.0 K
P1 = 8.7*10^-2 mmHg
P2 = 0.287 mmHg
?H = 56.1 KJ/mol
= 56100.0 J/mol
use:
ln(P2/P1) = (?H/R)*(1/T1 - 1/T2)
ln(0.287/8.7*10^-2) = (56100.0/8.314)*(1/298.0 - 1/T2)
1.1936 = 6747.6546*(1/298.0 - 1/T2)
T2 = 315 K
= (315-273) oC
= 42 oC
Answer: b
Only 1 question at a time please
.
PI1- (Sec- 11-8) The vapor pressure of the organie liquid toluene at 2.docx
1. PI1. (Sec. 11.8) The vapor pressure of the organie liquid toluene at 20.0 °C is 22 is the new
temperature, in °C, if the vapor pressure is increased 42 mmHg? (All",-40.6 kJ/mol) & (R-
8314 J / mol . K) mm H g What a) 24.2 °C b) 25.6 °C c) 28.1 °C d) 31.8 °C e) 39.5 °C
P12. (Sec. 11.8) The vapor pressure of the organic solid naphthalene at 25 °C is 0.087 mm Hg.
What is the new temperature, in °C, if the vapor pressure is carefully increased to 0.287
mmHg? AHvap 56.1 kJ/mol) & (R- 8.314J/mol K) a) 11.9°C b) 41.5 °C c) 23.4°C d) 35.4
°C e) 50.2 °c P13. (Sec. 11.8) The heat of vaporization (AHap)for water is 40.6 kJ/mol. How
much energy, in kJ, is needed to convert 55.5 grams of liquid water to steam at 100 °C? a) 67.1
kJ b) 98.1 kJ c) 202 kJ d) 76.9 kJ e) 125 kJ P14. (Sec. 11.9) On the phase diagram shown below,
segmentcorresponds to the conditions of temperature and pressure under which the solid and the
gas of the substance are in equilibriumm
Solution
11)
T1 = 20.0 oC
=(20.0 + 273)K
= 293.0 K
P1 = 22 mmHg
P2 = 42 mmHg
?H = 40.6 KJ/mol
= 40600.0 J/mol
use:
ln(P2/P1) = (?H/R)*(1/T1 - 1/T2)
ln(42.0/22) = (40600.0/8.314)*(1/293.0 - 1/T2)
2. 0.6466 = 4883.3293*(1/293.0 - 1/T2)
T2 = 305 K
= (305-273) oC
= 32 oC
Answer: d
12)
T1 = 25.0 oC
=(25.0 + 273)K
= 298.0 K
P1 = 8.7*10^-2 mmHg
P2 = 0.287 mmHg
?H = 56.1 KJ/mol
= 56100.0 J/mol
use:
ln(P2/P1) = (?H/R)*(1/T1 - 1/T2)
ln(0.287/8.7*10^-2) = (56100.0/8.314)*(1/298.0 - 1/T2)
1.1936 = 6747.6546*(1/298.0 - 1/T2)
T2 = 315 K
= (315-273) oC
= 42 oC
Answer: b
Only 1 question at a time please
