Consider the following data from a Heat of Vaporization experiment: (the volume

Question

Consider the following data from a Heat of Vaporization experiment: (the volume correction factor has not been applied -this is the raw data) Calculate the moles of air trapped in the cylinder. (Use the one low temperature point and assume Barometric pressure equals the pressure of only air since PH20 will be negligible.) Calculate the vapor pressure of water, P_water at the 59.8 degree C data point. First calculate P_air at 59.8 degree C using the moles of air. Then P_H20 = Barometric Pressure- P_air) Assuming the slope of the line resulting from a Claussius-Clapeyron plot (LnV.P. vs 1/T) of the data to be -5285 K, what is the value of delta H_vap in kJ/mol.

Explanation / Answer

a. moles of air trapped (n) = PV/RT

with,

P = (758.3/760) atm

V = 0.00465 L

R = gas constant

T = 273 + 2.9 = 275.9 K

we get,

moles of trapped air = (758.3/760)(0.00465)/0.08205)(275.9)

                                 = 2.05 x 10^-4 mol

b. Pair = nRT/V

            = 2.05 x 10^-4 x 0.08205 x (273 + 59.8)/(0.00698)

            = 0.802 atm

P(H2O) = 758.3 - 0.802 x 760 = 148.8 torr

c. slope = -dHvap/R

             = -5285

So,

dHvap = -5285 x 8.314/1000

           = -43.94 kJ/mol

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