Question

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization of benzene is 31.0 kJ/mol

Answer 1

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

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

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

Answer 2

Answer:

330.7mmHg

Explanation:

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