Final answer:
The boiling point of heptane at an external pressure of 0.614 atm is approximately 360 K. This can be calculated using the Clausius-Clapeyron equation with the known values for the normal boiling point and molar heat of vaporization of heptane.
Step-by-step explanation:
The boiling point of a liquid is the temperature at which its vapor pressure is equal to the external pressure. The Clausius-Clapeyron equation can be used to calculate the boiling point of a substance at a given external pressure. The equation is ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁), where P₂ is the vapor pressure at the given external pressure, P₁ is the vapor pressure at the normal boiling point, ΔHvap is the molar heat of vaporization, R is the ideal gas constant, T₂ is the boiling point at the given external pressure, and T₁ is the normal boiling point.
In this case, we want to find the boiling point of heptane (C₇H₁₆) when the external pressure is 0.614 atm. The normal boiling point of heptane is given as 371 K and the molar heat of vaporization is 34.7 kJ/mol. To solve for T₂, we can rearrange the Clausius-Clapeyron equation as T₂ = (ΔHvap/R)((1/T₁) - (1/T₂)) + 1/T₁, substitute the known values, and solve for T₂. Doing this calculation yields T₂ ≈ 360 K. Therefore, the boiling point of heptane at an external pressure of 0.614 atm is approximately 360 K.