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The vapor pressure of ethanol is 1.00 × 10² mmHg at 34.90°C. What is its vapor pressure at 58.05°C? (ΔHvap for ethanol is 39.3 kJ/mol.)

_____ mmHg

Answer :

Final answer:

The vapor pressure of ethanol at 58.05°C is approximately 1.02 × 10² mmHg.

Explanation:

To calculate the vapor pressure of ethanol at 58.05°C, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

Where:

  • P1 is the initial vapor pressure (1.00 × 10² mmHg)
  • P2 is the final vapor pressure (unknown)
  • ΔHvap is the enthalpy of vaporization for ethanol (39.3 kJ/mol)
  • R is the ideal gas constant (8.314 J/(mol·K))
  • T1 is the initial temperature (34.90°C + 273.15 K)
  • T2 is the final temperature (58.05°C + 273.15 K)

Let's plug in the values and solve for P2:

ln(P2/1.00 × 10²) = (39.3 * 10³)/(8.314) * (1/(34.90 + 273.15) - 1/(58.05 + 273.15))

Simplifying the equation:

ln(P2/1.00 × 10²) = 0.0147

Using the natural logarithm properties:

P2/1.00 × 10² = e(0.0147)

P2 = 1.00 × 10² * e(0.0147)

Calculating the value:

P2 ≈ 1.02 × 10² mmHg

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