Practice Exercise 11.18 Ethanol, with H ap = 40.5 k, morl has a higher dH a than

Question

Practice Exercise 11.18 Ethanol, with H ap = 40.5 k, morl has a higher dH a than methanol which is 35.2 k, mori due to its larger size and greater intermolecular forces. Because the ap for ethanol is larger, we would expect that for the two liquids to reach the same vapor pressure, the temperature for ethanol would have to be higher. If the vapor pressure is 0.0992 atm at 27.3 °C, what would the temperature be when the vapor pressure is 4.13 atm? The temperature will be K. The number of significant digits is set to 3; the tolerance is +/-196

Explanation / Answer

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(4.13/0.0992) = 35200/8.314*(1/(27.3+273) - 1/(T2))

T2 = -(ln(4.13/0.0992) /35200 * 8.314 - 1/(27.3+273) )^-1

T2 = 408.285 K

T2 = 408.285-273.15

T2 = 135.13 °C

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