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Please answer my chemistry question below which is abt finding heat of vaporization & normal boiling point?

The vapor pressure of nitrogen at several different temperatures is shown below. Use the data to find A) heat of vaporization & B) normal boiling point. Temperature (K), Pressure (torr)65 , 130.570 , 289.575 , 570.880 , 102885 , 1718*PLEASE GIVE ME THE COMPLETE SOLUTION & NOT JUST THE FINAL ANSWERS.

2 Answers

E
Everardo Feil

Feb 20, 2025

The vapor pressure of nitrogen at several different temperatures is shown below. Use the data to find A) heat of vaporization & B) normal boiling point. When the vapor pressure = atmospheric pressure, the liquid boils!Standard atmospheric pressure = 760 torrTemperature (K), Pressure (torr)65 , 130.570 , 289.575 , 570.880 , 102885 , 1718You need an equation for the graph of the data above, see web site below! https://eee.uci.edu/programs/gchem/05MANPTgasVPliq...ln P = (-1 Heat of vaporization ÷ R) (1/T) + CR = 8.3145 J/(mole * °K)T = temperature in °Kln P = (-1 Heat of vaporization ÷ 8.3145) (1/T) + CUse several values of T and P to determine the value of Heat of vaporizationln 130.5 = (-1 Heat of vaporization ÷ 8.3145) (1/65) + CC = ln 130.5 – (-1 Heat of vaporization ÷ 8.3145) (1/65)ln 570.8 = (-1 Hv ÷ 8.3145) (1/75) + ln 130.5 – (-1 Hv ÷ 8.3145) (1/65)Subtract ln 130.5 from both sidesln 570.8 – ln 130.5 = (-1 Hv ÷ 8.3145) (1/75) – (-1 Hv ÷ 8.3145) (1/65)ln 570.8 – ln 130.5 = Hv (-1 ÷ 8.3145) (1/75) – (-1 ÷ 8.3145)* (1/65)ln 570.8 – ln 130.5 = ( Hv /8.3145) * [(-1/75) – (-1/65)](6.347 – 4.871) 8.3145 = Hv [(-1/75) – (-1/65)]Hv = (1.476 * 8.3145) ÷ 0.00205Hv = 5986.44C = ln 130.5 – (-1 5986.44 ÷ 8.3145) (1/65)C = ln 130.5 – (-11.077)C = 15.95Then use the equation to determine the Temperature when P = 760 torr ln P = (-720) * (1/T) + 15.95P = 760ln 760 = (-720) * (1/T) + 15.95 ln 760 – 15.95 = (-720) * (1/T)(ln 760 – 15.95) ÷ (-720) = (1/T)T = 77.3 °Khttp://en.wikipedia.org/wiki/NitrogenBP = 77.36 °KWOW... Show More
E
Everardo Feil

Feb 20, 2025

This Site Might Help You.RE:Please answer my chemistry question below which is abt finding heat of vaporization & normal boiling point?The vapor pressure of nitrogen at several different temperatures is shown below. Use the data to find A) heat of vaporization & B) normal boiling point. Temperature (K), Pressure (torr)65 , 130.570 , 289.575 , 570.880 , 102885 , 1718*PLEASE GIVE ME THE COMPLETE SOLUTION & NOT JUST...... Show More

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