Question

1) Calculate the amount of energy (in kJ) necessary to convert 357 g of liquid water from 0*C to water vapor at 172*C. The molar heat of vaporization of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g*C, and for steam is 1.99 J/g*C.

2) The vapor pressure of ethanol is 1.00 x 10^2 mmHg at 34.90*C. What is its vapor pressure at 54.83*C? (deltaHvap for ethanol is 39.3 kJ/mol)

3) The vapor pressure of a liquid doubles when the temperature is raised from 85*C to 96*C. At what temperature will the vapor pressure be seven times the value at 85*C?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

1

Calculate the amount of energy (in Kj) necessary to convert 317g of liquid water from 0*c to water vapour at 172*c. The molar heat of vaporization of water is 40.79kj/mol. The specific heat for water is 4.184J/g/*c, and for steam is 1.99J/g/*c.

heat energy required to raise the temp of liquid water to 100 C = 317 * 4.184 * 100 = 132632 J =132.6 kJ

mole of water = 317 / 18 = 17.6 mol

heat required for vaporization = 17.6 * 40.79 = 717.9 kJ

heat required to raise the temp of water vapor to 172 C = 317 * 1.99 * 72 = 45420 J = 45.42 kJ

total heat required = 132.6 + 717.9 + 45.42 = 896 kJ

2

The vapor pressure of ethanol is 1.00 x 10^2 mmHg at 34.90 degree C. What is its vapor pressure at 60.78 degree C? (delta H vap for ethanol is 39.3 kJ/mol.)

For this kind of questions we use the clausius clapeyron equation

ln (P1 / P2) = ?Hvap / R x (1 / T2 - 1 / T1)

P1 = ???
P2 = 100 mmHg
?Hvap = 39.3 kJ = 39.3 x10^3 J/mole
R = 8.314 J/moleK
T2 = 34.9 C = 34.9+273.15 K = 308.05 K
T1 = 60.78 C = 60.78 +273.15 K = 333.93 K

ln (P1 / 100 mmHg) = [(39300 J/mole) / (8.314 J/moleK)] x (1 / 308.05 K - 1 / 333.93 K)

ln (P1 / 100 mmHg) = 4227K x (2.52 x10^-4 / K)

ln (P1 / 100 mmHg) = 1.189

e^ln (P1 / 100 mmHg) = e^(1.549)

P1 = 100 mmHg x 3.284 = 328.45 mmHg = 328 mm Hg

3

The vapor pressure of a liquid doubles when the temperature is raised from

ln2 = dH/R (1/78-1/88)
dH/R = ln2/ (1/78-1/88)

ln4 = dH/R (1/78-1/T)
1/78- 2*(1/78-1/88)) = 1/T
T= 100.941 C