Final answer:
The vapor pressure of substance X at -82.0°C is 0.048 atm.
Step-by-step explanation:
To calculate the vapor pressure of substance X at -82.0°C, we can use the Clausius-Clapeyron equation. The equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature and the enthalpy of vaporization.
The Clausius-Clapeyron equation is given by ln(P₂/P₁) = -ΔHvap/R * (1/T₂ - 1/T₁), where P₁ is the vapor pressure at temperature T₁, P₂ is the vapor pressure at temperature T₂, ΔHvap is the enthalpy of vaporization, and R is the gas constant.
Using the given values, where T₁ = -34.0°C = 239.15 K and P₁ = 1 atm, and solving for P₂ with T₂ = -82.0°C = 191.15 K, we can calculate the vapor pressure. Plugging the values into the equation gives us ln(P₂/1) = -24.0 kJ/mol / (8.314 J/mol K) * (1/191.15 K - 1/239.15 K).
Solving for P₂ gives us P₂ = 0.048 atm.