High School

What is the vapor pressure of ethanol (in mmHg) at 62.1 °C if its vapor pressure is 400.0 mmHg at 63.5 °C?

Given:
- \(\Delta H_{\text{vap}} = 39.3 \, \text{kJ/mol}\)
- \(R = 8.314 \, \text{J/K} \cdot \text{mol}\)

Answer :

Final Answer:

The vapor pressure of ethanol at 62.1 °C is about 382.75 mmHg, calculated using the Clausius-Clapeyron equation with given data and constants.

Explanation:

To calculate the vapor pressure of ethanol at 62.1 °C using the Clausius-Clapeyron equation, we can use the following formula:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Where:

P1 = Vapor pressure at temperature T1 (63.5 °C)

P2 = Vapor pressure at the desired temperature (62.1 °C)

ΔHvap = Enthalpy of vaporization

R = Gas constant (8.314 J/K•mol)

T1 = Temperature in Kelvin corresponding to T1

T2 = Temperature in Kelvin corresponding to T2

First, let's convert the temperatures to Kelvin:

T1 = 63.5 + 273.15 = 336.65 K

T2 = 62.1 + 273.15 = 335.25 K

Now, plug in the values and solve for ln(P2/P1):

ln(P2/400.0) = -(39300 J/mol) / (8.314 J/K•mol) * (1/335.25 K - 1/336.65 K)

ln(P2/400.0) = -4727.48 * (0.002985 - 0.002975)

ln(P2/400.0) = -4727.48 * 0.000010

ln(P2/400.0) = -0.0472748

Now, solve for P2:

P2/400.0 = [tex]e^(-0.0[/tex]472748)

P2 = 400.0 * [tex]e^(-0.0[/tex]472748)

P2 ≈ 382.75 mmHg

So, the vapor pressure of ethanol at 62.1 °C is approximately 382.75 mmHg.

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