Cyclohexanol has a vapor pressure of 10.0 mmHg at 56.0 C and 100.0 mmHg at 103.7

Question

Cyclohexanol has a vapor pressure of 10.0 mmHg at 56.0 C and 100.0 mmHg at 103.7 C. Calculater its Hvap.

The answer was given at 49.7kj/mol, I just need the steps to follow to get me to the answer.
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Explanation / Answer

Originally the Clausius Clapeyron relations states that the derivative of vapor pressure with respect to temperature is equal to the ratio of heat of vaporization to change of volume in vaporization: dp/dT = ?Hv / ?Vv Approximating that the volume change is equal to the gaseous molar volume, given by ideal gas law: ?Vv = R · T / p leads to the more convenient from of the relation: (1/p)· dp/dT = ?Hv / (R·T²) Separation of variables an integration gives the integrated form of the relation: ? (1/p)dp = ? ?Hv / (R·T²) dT => lnp = - ?Hv/(R·T) + C (c is the constant of integration) Here we got the vapor pressures at two temperatures: lnp1 = - ?Hv/(R·T1) + C lnp2 = - ?Hv/(R·T2) + C => lnp2 - lnp1= - ?Hv/(R·T2) + - ?Hv/(R·T1) ln(p2/p1) = ?Hv/R · [(1/T1) - (1/T2)] Solve for the heat of vaporization: => ?Hv = R · ln(p2/p1) / [(1/T1) - (1/T2)] = 8.31772J/molK · ln(100mmHg / 10mHg) / [(1 / 329.15K) - (1/376.85K)] = 49784.5J/mol = 49.8kJ/mol

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