Question

Use the data in the table below to calculate the heat of vaporization (AHvap) in kJ/mol of pinene.

Vapor Pressure
(torr)
760
515
340
218
135


Temperature
(K)
429
415
401
387
373
kJ/mol
Use the value of AHyap determined in Part 1 to calculate the vapor pressure of pinene (in torr) at room temperature (23°C)
760
torr

Answer 1

• 41 kJ/mol
• 4 torr

Step-by-step explanation:

Given pinene has a (temperature, vapor pressure) relation (K, torr) = {(373, 135), (429, 760)}, you want the heat of vaporization in kJ/mol and the vapor pressure at room temperature (23 °C).

Clausius–Clapeyron Equation

The Clausius–Clapeyron equation can be used to find the heat of vaporization:

`ln(P)=-\frac{\Delta H_{\text{vap}}}{R}\left((1)/(T)\right)+C`

Solving for ∆H, we find ...

`\Delta H_{\text{vap}}=-\frac{R\cdot\ln{(P_1)/(P_2)}}{(1)/(T_1)-(1)/(T_2)}\\\\\\\Delta H_{\text{vap}}=-\frac{8.314\cdot\ln{(760)/(135)}}{(1)/(429)-(1)/(373)}\approx 41052.8`

The heat of vaporization of pinene is about 41 kJ/mol.

Vapor pressure

Rearranging the above equation to give P1, we have ...

`\ln{(P_1)/(P_2)}=-\frac{\Delta H_\text{vap}}{R}\left((1)/(T_1)-(1)/(T_2)\right) \\\\\\P_1=P_2\cdot e^{-\frac{\Delta H_\text{vap}}{R}\left(T_1^(-1)-T_2^(-1))}`

Using the same P2 and T2 as above, we find the vapor pressure at room temperature (296.15 K) to be ...

P1 ≈ 4.349 . . . . . torr

The vapor pressure of pinene at room temperature is about 4 torr.