Chapter 10, Problem 10.63QP

(a)

Interpretation Introduction

Interpretation: The heat of vaporization of the given pinene is to be calculated. The vapor pressure of the pinene at room temperature is to be calculated.

Concept introduction: Vapor pressure is defined as the pressure applied by the molecule in the vapor state is in equilibrium with the solid or liquid state of the same molecule in a closed space.

The factors affect the vapor pressure of a liquid are temperature, surface area and type of a liquid.

If the molecule in the liquid state is more volatile means it will be evaporates easily so it vapors pressure is higher than the other molecule in the liquid state.

To determine: The heat of vaporization of pinene using the given table.

Answer to Problem 10.63QP

Solution

The heat of vaporization of pinene is $\underset{\_}{41KJ/mol}$.

Explanation of Solution

Explanation

Given

The value of $R$ is $8.314\text{J/mol}\text{.K}$.

The given structure is below.

Figure 1

In the above Figure, the carbon-skeleton structure of pinene is depicted.

The clausius-clapeyron equation states the relation between vapor pressure and the absolute temperature.

The clausius-clapeyron equation is,

$\ln \left(\frac{P_{\text{vap},T_1}}{P_{\text{vap},T_2}}\right)=\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ (1)

Where,

• $P_{\text{vap},T_1}$ is vapor pressure at temperature $T_1$.
• $P_{\text{vap},T_2}$ is vapor pressure at temperature $T_2$.
• $\Delta H_{\text{vap}}$ is heat of vaporization.
• $R$ is universal gas constant.
• $T_1$ is the first absolute temperature.
• $T_2$ is the second absolute temperature.

The table of different values of vapor pressure at different temperature of pinene is given below.

$\begin{array}{cc}\text{Vapor}\text{pressure(torr)}& \text{Temperature(K)}\\ 760& 429\\ 515& 415\\ 340& 401\\ 218& 387\\ 135& 373\end{array}$

Table 1

The heat of vaporization of pinene is calculated by the above equation (1).

Substitute the any two values of vapor pressure at the corresponding temperature from the table in the equation (1).

$\begin{array}{c}\ln \left(\frac{760\text{torr}}{515\text{torr}}\right)=\frac{\Delta H_{\text{vap}}}{8.314\text{J/mol}\text{.K}}\left(\frac{1}{415\text{K}}-\frac{1}{429\text{K}}\right)\\ \ln \left(1.4757\right)=\frac{\Delta H_{\text{vap}}}{8.314\text{J/mol}\text{.K}}\left(\frac{14}{178,035}\right)\text{K}\\ \Delta H_{\text{vap}}=\frac{0.389\times 8.314\text{J/mol}\times \text{178,035}}{14}\\ \Delta H_{\text{vap}}=\underset{\_}{41KJ/mol}\end{array}$

Therefore, heat of vaporization of pinene is $\underset{\_}{41KJ/mol}$.

(b)

Interpretation Introduction

To determine: The vapor pressure of the pinene at room temperature.

Answer to Problem 10.63QP

Solution

The vapor pressure of the pinene at room temperature $\left(\text{23}\text{°C}\right)$ is $\underset{\_}{4.27torr}$.

Explanation of Solution

Explanation

Given

The value of $R$ is $8.314\text{J/mol}\text{.K}$.

The room temperature is $\text{23}\text{°C}$.

The vapor pressure of the pinene at room temperature is calculated by equation (1) given in part (a).

The heat of vaporization of pinene is $41\text{KJ/mol}$ as calculated above in part (a).

The room temperature in Kelvin is $\text{296}\text{K}$.

Substitute the given value of $R$, room temperature and any value of vapor pressure at the corresponding temperature from the table given in part (a) in the above equation (1).

$\begin{array}{c}\ln \left(\frac{760\text{torr}}{P_{{\text{vap,T}}_2}}\right)=\frac{41\times {10}^3\text{J/mol}}{8.314\text{J/mol}\text{.K}}\left(\frac{1}{296}-\frac{1}{429}\right)\text{K}\\ \text{ln}\left(\frac{760\text{torr}}{P_{{\text{vap,T}}_2}}\right)=4946.83\times \left(\frac{133}{126984}\right)\\ \frac{7\text{60}\text{tor}r}{P_{{\text{vap,T}}_2}}=e^{5.18}\\ P_{{\text{vap,T}}_2}=\underset{\_}{4.27torr}\end{array}$

Therefore, the vapor pressure of the pinene at room temperature $\left(\text{23}\text{°C}\right)$ is $\underset{\_}{4.27torr}$.

Conclusion

1. a) The heat of vaporization of pinene is $\underset{\_}{41KJ/mol}$.
2. b) The vapor pressure of the pinene at room temperature $\left(\text{23}\text{°C}\right)$ is $\underset{\_}{4.27torr}$.