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------------------------------------------------ The vapor pressure of ethanol is 100 mmHg at 34.9 ∘C. What is its vapor pressure at 65.0 ∘C? (ΔH_vap for ethanol is 39.3 kJ/mol.)

The Clausius-Clapeyron equation establishes a connection between vapor pressure and temperature. It links the change in vapor pressure with temperature to the enthalpy of vaporization. The formula is represented as:

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

P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively.

ΔHvap stands for the enthalpy of vaporization.

R is the universal gas constant, which is 8.314 J/(mol·K).

Using the provided information:

P1 is 100 mmHg at T1, which equals 34.9°C or 308.05 K.

ΔHvap is 39.3 kJ/mol, which can also be stated as 39300 J/mol.

T2 is 65.0°C, converting to 338.15 K.

Substituting these values into our equation, we have:

ln(P2/100) = (-39300 J/mol / 8.314 J/(mol·K)) * (1/338.15 K - 1/308.05 K)

From this, we can determine P2 to be about 199.6 mmHg.

Therefore, it's evident that the vapor pressure of ethanol at 65.0°C is roughly 199.6 mmHg. This relationship showcases how an increase in temperature will lead to a rise in vapor pressure.

Learn more about Vapor pressure

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