Question

4. Calculate the boiling point of water at the top vaporization of Mt. Whitney, alt-4,421 meters. The normal heat of altitude can be calculated from Pz)-760 x emRT tor the altitude above the earth's surface of water is 540 calories per gram. You should recognize that the atmospheric pressure at any while R and T have their usual meanings the molar mass of air, g. 981 mw andOs Nornal heort of Vapon 2aen of 20: S40 Calor es pex P (4421 )(1000 M (9.01 Ms) de a earth's surfac ?433?0010 Ra (am)/? m: malar mass of air ospheric pressure 2 = 4,42/ meters Molar Mass odny aur The hon (-2890)(TB)( 12)/RT torr : 28 91 81ml oner the belm 120 wll bail @ a lamp less than 100'c

Answer 1

Ans.

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Step 1: Atmospheric pressure (Pz) at an altitude is given by - -mgh RT Pz= 760 torr x e - Equation 1 Where m = Molar mass of Air g = Acceleration due to gravity = h= Height RUniversal gas constant T = Temperature in kelvin 0.029 kg mol 9.807 m s 4421 m -2 8.314 J mol1 K F 298.15 K Putting the values in equation 1 -0.0290 kg mol x 9.807 m sx 4421 m 8.31446 J mol-1 К-1 x 298.15 K Pz760 torr x e -1255.76 kg m2s2 mol1 2478.96 Jmol Pz760 torr x e 1255.76 J mol -0.507 = 760 torr x e PZ = 760 torr xe 2478.96 J mol Pz760 torr x 0.603457.9 torr

# Step 2: Given –

Normal heat of vaporization, dHvap = 540 cal/ g

                                                            = (540 x 4.184 J) / [ (1 g / 18.015 g mol^-1)]

                                                            = 2259.36 J / (0.05551 mol^-1)

                                                            = 40701.86 J mol^-1

                                                            = 40.70 kJ mol^-1

            Normal temperature = 25.0⁰C

            Normal pressure = 1 atm

AHvap Clausius-Clapeyron Equation In P1 T1 T2 Where P1 vapor presure at temperature T1 P2 vapor presure at temperature T2 T1 and T2 = temperature (K) AHvap enthalpy of vaporization R-Uni, gas constant = 0.008314 kJ mol-"K-1 Given, P1= 760 mmHg = 1.0000 atm T1100 °c- 373.15 K ΔHvap = 40.70 kJ mol-1 P2= 457.9 mmHg= 0.6025 atm Putting the values in above expression 40.70 kJ mol 0.60 atm 2.303 log 1.00 atm 0.008314 kJ molK373.15 K T2 0.50666761 4895.357229 0.00010350 K1 0.002679887 K1 -_ T2 Or, Or, 0.002679887 K-0.00278339 K Hence, T2= 359.27 K 86.12 °c 0.002783387 K

# Therefore, the BP of water at specified altitude = 86.12⁰C