A certain substance has a heat of vaporization of 34.65 kj/mol. At what kelvin temperature will the vapor pressure be 3.50 times higher than it was at 331 k?
Use the Clausius-Clapeyron equation.
I would make up a vapor pressure for p1 and multiply that by 3.5 for p2; alternatively you can use p1 as p1 and p2 = 3.5p1.
To solve this problem, we need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and heat of vaporization.
The Clausius-Clapeyron equation is given by:
ln(P₂/P₁) = -(ΔHvap/R) * (1/T₂ - 1/T₁)
Where:
P₁ and P₂ are the initial and final vapor pressures, respectively.
ΔHvap is the heat of vaporization of the substance.
R is the ideal gas constant (8.314 J/(mol·K)).
T₁ and T₂ are the initial and final temperature, respectively, in Kelvin.
In this case, we are given:
P₁ = vapor pressure at 331 K
P₂ = 3.50 * P₁ (since it is 3.50 times higher)
ΔHvap = 34.65 kJ/mol (convert to J/mol: 34.65 kJ * 1000 = 34,650 J/mol)
T₁ = 331 K
T₂ = the temperature we want to find (in Kelvin)
Substituting these values into the equation, we have:
ln(3.50 * P₁/P₁) = -(34,650 J/mol / 8.314 J/(mol·K)) * (1/T₂ - 1/331 K)
Simplifying:
ln(3.50) = -4173.02 * (1/T₂ - 1/331 K)
Now, rearrange the equation to solve for 1/T₂:
1/T₂ = (ln(3.50) / -4173.02) + 1/331 K
1/T₂ = 0.0003778 + 0.00303
1/T₂ = 0.0034078
Now, solve for T₂:
T₂ = 1 / 0.0034078
T₂ ≈ 293.64 K
Therefore, at approximately 293.64 K, the vapor pressure will be 3.50 times higher than it was at 331 K.