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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.
 Fall '08
 dockery

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