Question

Ethanol has an enthalpy of vaporization of 42.3 kJ/mol. The compound has a vapor pressure of 1.00 atm at 78.3 degree C. At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K middot mol) -83.8 degree C -24.4 degree C 62.6 degree C 73.0 degree C 78.0 degree C A liquid has an enthalpy of vaporization of 30.4 kJ/mol. At 269 K is has a vapor pressure of 102 mmHg. What is the normal boiling point of this liquid? (R = 8.314 J/(K middot mol)) 287 K 316 K 269 K 253 K. 234 K Sulfur dioxide has a vapor pressure of 462.7 mm Hg at -21.0 degree C and a vapor pressure of 140.5 mm Hg at -44.0 degree C. What is the enthalpy of vaporization of sulfur dioxide? (R = 8.314 J/K middot mol) 0.398 kJ/mol 6.33 kJ/mol 14.0 kJ/mol 24.9 kJ/mol Mount Everest rises to a height of 8.850 times 10^3 m above sea level. At this height, the atmospheric pressure is 231 mm Hg. At what temperature (in degree C) does water boil at the summit of Mount Everest? The vapore pressure of water at 373 K is 760.0 mm Hg. (delta_vap H_2O = 40.7 kJ/mol and R = 8.314 J/K middot mol) 4.07 degree C 69.0 degree C 72 degree C 87 degree C 364 degree C

Can you show the work for these problems?

Answer 1

Ans. 30. Vapor pressure for normal boiling = 1.00 atm = 760 mmHg

Putting the values in equation 1-

ln (760 mmHg/ 102 mmHg) = [30400 J mol^-1 / 8.314 J K^-1mol^-1] [(1/269K) – (1/T2)]

Or,       2.00834562 = 3656.48 K [0.00372 K^‑1 – (1/T2)]

Or,       (2.00834562 / 3656.48 K) = 0.00372 K^‑1 – (1/T2)

Or,       0.0005493 K^-1 – 0.00372 K^-1 = – (1/T2)

Or,       T2 = 1/ (0.0031707 K^-1) = 315.38 K

Hence, the normal boiling point (Temperature at which vapor pressure = 1.0 atm) = 315.38 K

            Note: Using 0⁰C = 273.00 K may give a value much closer to 316 K

Ans. 31. Putting the values in equation 1

ln (462.7 mmHg/ 140.5 mmHg) = [dH / 8.314 J K^-1mol^-1] [(1/ 229.15K) – (1/ 252.15K)]

Or,       1.1918714071= [dH / 8.314 J K^-1mol^-1] x (0.004364 K^‑1 – 0.003966 K^-1)

Or,       1.1918714071= [dH / 8.314 J K^-1mol^-1] x (0.000398 K^-1)

Or,       1.1918714071 / 0.000398 K^-1 = dH / 8.314 J K^-1mol^-1

Or,       (2994.65 K^-1) x 8.314 J K^-1mol^-1= dH

Or,       dH = 24897.53 J mol^-1 = 24.9 kJ mol^-1                                   ; [1 J = 10^-3 kJ]

Hence, heat of vaporization = 24.9 kJ mol^-1

Ans. 32. Putting the values in equation 1

ln (231 mmHg/ 760.0 mmHg) = [40700 J mol^-1 / 8.314 J K^-1mol^-1] [(1/ 373 K) – (1/ T2)]

Or,       - 1.19= [4895.36 K^-1] x (0.002681 K^‑1 – 1/T2)

Or,       - 1.19/ 4895.36 K^-1 =0.002681 K^‑1 – 1/T2

Or,       - 0.000243 K^-1 - 0.002681 K^‑1 = – 1/T2

Or,       T2 = 1 / (0.002924 K^-1) = 341.99 K                 

Hence, T2 = 341.99 K = 68.99⁰C     

Note and Cautions:

1. Always write the values of T1 and P1 at their respective places. So, do it for T2 and P2.

However, it does not matter which set (of P1 and T1 ; or P2 and T2) of T and P you take as first and second conditions.

2. Temperature shall always be in K.

3. Note the sign. ln of a value less than 1.0 (say, 0.998) is a negative integer and that of greater than 1.0 is a positive integer.