Asked by Camaro Lover
The normal boiling point of liquid ethanol is 351 K. Assuming that its molar heat of vaporization is constant at 37.5 kJ/mol, the boiling point of C2H5OH when the external pressure is 1.35 atm is K.
Answers
Answered by
DrBob222
Use the Clausius-Clapeyron equation.
You know p = 1 atm when T = 351K. Solve for p2 when T = 1.35 atm.
You know p = 1 atm when T = 351K. Solve for p2 when T = 1.35 atm.
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