Problem Breakdown

The liquid boils normally at 78.3°C (351.45 K) under
760 Torr, where its vapor pressure equals atmospheric pressure.
To make the liquid boil at a lower temperature of
25°C (298.15 K), the external pressure must be reduced.

Given data:

• Normal boiling point, T₁ = 78.3°C = 351.45 K
• Required temperature, T₂ = 25°C = 298.15 K
• Normal pressure, P₁ = 760 Torr
• ΔH_vap = 39 kJ/mol = 39,000 J/mol
• R = 8.314 J·K⁻¹·mol⁻¹

Step-by-Step Solution

The Clausius–Clapeyron equation is:

ln(P₂/P₁) =
−(ΔH_vap/R)
(1/T₂ − 1/T₁)

Substitute known values:

• P₁ = 760 Torr
• T₁ = 351.45 K
• T₂ = 298.15 K

Calculate temperature term:

1/T₂ − 1/T₁ =
0.003353 − 0.002846 =
0.000507 K⁻¹

Calculate ΔH_vap/R:

39,000 / 8.314 ≈ 4,690 K

Substitute into equation:

ln(P₂/760) = −4,690 × 0.000507 ≈ −2.378

Exponentiating both sides:

P₂/760 = e^−2.378 ≈ 0.1125

Therefore:

P₂ ≈ 85.5 Torr

Why This Approach Works

The Clausius–Clapeyron equation assumes constant ΔH_vap and ideal gas behavior,
which is a valid approximation for many liquids near their boiling points.
Lower temperatures reduce vapor pressure exponentially, requiring a
vacuum (reduced pressure) to induce boiling.

For ethanol-like liquids with a normal boiling point near 78°C,
a pressure of approximately 85.5 Torr allows boiling at room temperature.

Common Mistakes in Exam Options

• Using °C instead of Kelvin: Leads to large numerical errors.
• Not converting kJ to J: Forgetting ×1000 gives unrealistically high pressures.
• Swapping T₁ and T₂: Predicts pressure greater than 760 Torr.
• Wrong gas constant: Using 0.0821 without unit conversion distorts results by ~10×.

When calculated correctly, the pressure required remains
85.5 Torr, accurate to one decimal place.