Final answer:
Heptane's boiling point decreases as the external pressure is reduced; however, without sufficient information to apply the Clausius-Clapeyron equation, a numerical answer cannot be given.
Step-by-step explanation:
The boiling point of a liquid changes with the external pressure. For heptane (C7H16) with a normal boiling point of 371 K and a molar heat of vaporization of 34.7 kJ/mol, we would use the Clausius-Clapeyron equation to estimate its boiling point at a different external pressure.
However, since the question does not provide enough information to apply the Clausius-Clapeyron equation, such as the initial vapor pressure or the need to make a linear approximation over a temperature range, it isn't possible to give a precise answer without additional data or assumptions. In such cases, the boiling point would decrease as the external pressure is reduced from 1 atm to 0.788 atm, but a numerical answer cannot be provided.
The boiling point of a liquid changes with the external pressure. To find the boiling point of heptane (C₇H₁₆) when the external pressure is 0.788 atm, we can use the Clausius-Clapeyron equation:
ln(P₁/P₂) = (ΔHvap/R) * (1/T₁ - 1/T₂)
We know that the normal boiling point of heptane (T₁) is 371 K and the molar heat of vaporization (ΔHvap) is 34.7 kJ/mol. Rearranging the equation and solving for T₂ (the boiling point at 0.788 atm), we get:
T₂ = (ΔHvap/R) * ((1/T₁) - ln(P₂/P₁))
Substituting the given values, we can calculate the boiling point:
T₂ = (34.7 kJ/mol / 8.314 J/K·mol) * ((1/371 K) - ln(0.788 atm / 1 atm))
T₂ ≈ 342 K