Vapor%20Pressure%20Answers

Vapor%20Pressure%20Answers - 0.000280203 = (1 / T2...

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Vapor Pressure Calculations 1. Given that a substance has a vapor pressure of 85 torr at 15 Celsius, solve for the vapor pressure at -15 Celsius. The H vap = 65 kJ/mol for this substance. ln (P2 / 85) = (−65000 / 8.314) * (1 / 258.15 – 1 / 288.15) ln (P2 / 85) = (−65000 / 8.314) * 0.000403302 ln (P2 / 85) = -3.153 (P2 / 85) = 0.04272 P2 = 3.6 torr 2. In the above problem, solve the temperature (in Celsius) where the vapor pressure would be 350 torr. ln (350 / 85) = (−65000 / 8.314) * (1 / T2 – 1 / 288.15) 1.415 = −7818.14 * (1 / T2 – 0.0034704) −0.000180989 = (1 / T2 – 0.0034704) 0.0032894 = 1 / T2 T2 = 304.00 Kelvin 30.86 Celsius 31 Celsius 3. In problem 1, find the boiling point of the substance (in Celsius) at atmospheric pressure. Note, atmospheric pressure is equal to 760 torr. ln (760 / 85) = (−65000 / 8.314) * (1 / T2 – 1 / 288.15) 2.191 = −7818.14 * (1 / T2 – 0.0034704)
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Unformatted text preview: 0.000280203 = (1 / T2 0.0034704) 0.0031902 = 1 / T2 T2 = 313.46 Kelvin 40.31 Celsius 40. Celsius 4. Given a new substance that has a vapor pressure of 0.15 atm at 20 Celsius, and the vapor pressure doubles when you heat it 10 degrees Celsius, solve for the H vap of this substance. T1 = 20 Celsius = 293.15 P1 = 0.15 atm T2 = 30 Celsius = 303.15 P2 = 0.30 atm ln (0.30 / 0.15) = ( H vap / 8.314) * (1 / 303.15 1 / 293.15) 0.69315 = ( H vap / 8.314) * (0.00011253) 6159.915 = ( H vap / 8.314) 51213.53 J / mol = H vap = 50 kJ/mol Note, this problem is still solvable if the vapor pressure (0.15 atm) was not given in the problem. You could have used P2 = 2 * P1 (since the vapor pressure doubled) and be left with ln(2 * P1 / P1) in the equation, becoming ln(2). We used actual values and had ln(2)....
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This note was uploaded on 11/29/2011 for the course CHEM 1212 taught by Professor Dockery during the Fall '08 term at Kennesaw.

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