Question

A certain substance has a heat of vaporization of 64.08 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.50 times higher than it was at 355 K?

Answer 1

Answer:

$T_2 =388.50K$

Explanation:

Given

$p_1 = ??$ -- Initial vapour pressure

$p_2 = 6.50p_1$ --- Final vapour pressure

$T_1 = 355K$ ---- Initial temperature

$T_2 = ??$ --- Final temperature

$\triangle _{vap}H = 64.08kJmol$ --- Enthalpy of vaporization

Required

Calculate T2

To do this, we make use of Clausius-Clapeyron equation.

Which states that:

$ln(\frac{p_2}{p_1}) = \frac{\triangle _{vap}H}{R}(\frac{1}{T_1} - \frac{1}{T_2})$

Where:

$R = 8.314 J.K^{-1}mol^{-1}$ --- Universal Gas constant

$\triangle _{vap}H = 64.08kJmol$

$\triangle _{vap}H = 64080\ kJmol$

$ln(\frac{p_2}{p_1}) = \frac{\triangle _{vap}H}{R}(\frac{1}{T_1} - \frac{1}{T_2})$ becomes

$ln(\frac{6.50p_1}{p_1}) = \frac{64080}{8.314}(\frac{1}{355} - \frac{1}{T_2})$

$ln(\frac{6.50p_1}{p_1}) = 7707.48(\frac{1}{355} - \frac{1}{T_2})$

$ln(6.50) = 7707.48(\frac{1}{355} - \frac{1}{T_2})$

$1.872 = 7707.48(\frac{1}{355} - \frac{1}{T_2})$

Take LCM

$1.872 = 7707.48(\frac{T_2 - 355}{355T_2})$

$1.872 = 7707.48*\frac{T_2 - 355}{355T_2}$

$1.872 = \frac{7707.48*(T_2 - 355)}{355T_2}$

Cross Multiply

$355T_2*1.872 = 7707.48*(T_2 - 355)$

$664.56T_2 = 7707.48T_2 - 2736155.4$

Collect Like Terms

$664.56T_2 - 7707.48T_2 =- 2736155.4$

$-7042.92T_2 =- 2736155.4$

Make T2 the subject

$T_2 =\frac{- 2736155.4}{-7042.92}$

$T_2 =388.497299416$

$T_2 =388.50K$

The final temperature is 388.50K