A certain substance has a heat of vaporization of 26.49 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 285 K?
Use the Clausius-Clapeyron equation.
To solve this question, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures to its heat of vaporization.
The Clausius-Clapeyron equation is given by:
ln(P₂/P₁) = -ΔH_vap/R * (1/T₂ - 1/T₁)
Where P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively, ΔH_vap is the heat of vaporization, R is the ideal gas constant (8.314 J/(mol·K)), and ln denotes the natural logarithm.
In this case, we know the vapor pressure at T₁ = 285 K, which we'll call P₁, and we want to find the temperature T₂ at which the vapor pressure is 5.00 times higher than P₁. So we can write P₂ = 5 * P₁.
Now, let's substitute these values into the Clausius-Clapeyron equation and solve for T₂:
ln(5 * P₁/P₁) = -ΔH_vap/R * (1/T₂ - 1/285)
Since P₁/P₁ equals 1, the equation simplifies to:
ln(5) = -ΔH_vap/R * (1/T₂ - 1/285)
Now, let's rearrange the equation to solve for T₂:
1/T₂ - 1/285 = (ln(5) * R) / -ΔH_vap
Combining the numerators:
1/T₂ - 1/285 = -ln(5) * R / ΔH_vap
Now, let's simplify the equation further:
1/T₂ = (-ln(5) * R / ΔH_vap) + 1/285
1/T₂ = (-ln(5) * 8.314 J/(mol·K) / 26490 J/mol) + 1/285
Now, let's solve for T₂:
T₂ = 1 / [(ln(5) * 8.314 J/(mol·K) / 26490 J/mol) + 1/285]
Calculating this value using a calculator, we find that T₂ is approximately 388.5 K. Therefore, at a temperature of approximately 388.5 K, the vapor pressure will be 5.00 times higher than it was at 285 K.