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Answer :
The given temperature of ethanol is 28 °C, and its boiling point is 78 °C. Thus, the temperature given is less than its boiling point, which means that the ethanol is in the liquid state, not in the gaseous state. The answer is option B. 0.111atm.
This means that the vapor pressure is the saturated vapor pressure of ethanol at 28 °C. The Clausius-Clapeyron equation is used to calculate the saturated vapor pressure. The equation is given as:
log P2/P1 = ΔHvap/R × (1/T1 - 1/T2)
where ΔHvap is the heat of vaporization, R is the gas constant, T1 is the boiling point of the liquid, T2 is the temperature for which the saturated vapor pressure is to be calculated, P1 is the vapor pressure at T1, and P2 is the vapor pressure at T2.The values are given as follows:
ΔHvap = 38.6 kJ/molR
= 8.314 J/mol.
KT1 = 78 °C + 273.15
= 351.15 K (boiling point of ethanol)
T2 = 28 °C + 273.15
= 301.15 K (temperature given)
P1 = atmospheric pressure (because the boiling point of ethanol is above the atmospheric pressure)P2 = ?log P2/atm atmospheric pressure/atm = 0.111atmapproximately.So, the answer is option B. 0.111atm.
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