Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
[tex]ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}][/tex]
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
[tex]ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg[/tex]
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg
330.7mmHg
The relationship between pressure and temperature, to estimate vapor pressure at any temperature or to estimate the heat of vaporization of phase transition from the vapor pressure at two temperatures, is given by the Clausius-Clapeyron equation as follows;
ln (P₂ / P₁) = (- ΔH / R) [ ( 1 / T₂ ) - ( 1 / T₁) ] ------------------- (i)
Where;
P₂ and P₁ are the final and initial pressures respectively
T₂ and T₁ are the final and initial temperatures respectively
R = Universal Gas constant
ΔH is the heat of vaporization
Given, from question;
P₁ = 40.1mmHg
T₁ = 7.6°C = (273 + 7.6)K = 280.6K
T₂ = 60.6°C = (273 + 60.6)K = 333.6K
ΔH = 31.0kJ/mol = 31000 J/mol
Known constant;
R = 8.314 J/mol.K
Substitute these values into equation (i)
ln (P₂ / 40.1) = (- 31000 / 8.314) [ (1 / 333.6) - ( 1 / 280.6)]
ln (P₂ / 40.1) = (-3728.65) [ (0.002998) - (0.003564)]
ln (P₂ / 40.1) = (-3728.65) [-0.000566]
ln (P₂ / 40.1) = 2.11
Solve for P₂ by taking the inverse ln of both sides;
P₂ / 40.1 = e²°¹¹
P₂ / 40.1 = 8.248
P₂ = 8.248 x 40.1
P₂ = 330.7mmHg
Therefore, the vapor pressure at 60.6°C is 330.7mmHg
