Chapter 12, Problem 12.129P

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of methyl salicylate at $25\text{°C}$ is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to find the vapor pressure at one temperature when vapor pressure at another temperature and heat of enthalpy is given. It determines the change in vapor pressure with a change in temperature. The expression of the Clausius-Clapeyron equation is as follows:

$\ln \left(\frac{P_2}{P_1}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$        (1)

Here,

$P_1\text{ and }{P}_2$ are the vapor pressure.

$T_1\text{ and }{T}_2$ are the temperature.

$R$ is the gas constant.

$\Delta H_{\text{vap}}$ is the heat of vaporization.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.2$        (2)

Answer to Problem 12.129P

The vapor pressure of methyl salicylate at $25\text{°C}$ is $0.13\text{torr}$.

Explanation of Solution

Substitute $95.3°\text{C}$ in equation (2) to calculate the temperature $T_1$ in Kelvin as follows:

$\begin{array}{c}{T}_1\left(\text{K}\right)=95.3°\text{C}+273.2\\ =368.5\text{ K}\end{array}$

Substitute $54.3°\text{C}$ in equation (2) to calculate the temperature $T_2$ in Kelvin as follows:

$\begin{array}{c}{T}_2\left(\text{K}\right)=54.3°\text{C}+273.2\\ =327.5\text{ K}\end{array}$

Rearrange the equation (1) to calculate $\Delta H_{\text{vap}}$.

$\Delta H_{\text{vap}}=-\frac{\left(R\right)\ln \left(\frac{P_2}{P_1}\right)}{\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$        (3)

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $368.5\text{ K}$ for $T_1$, $327.5\text{ K}$ for $T_2$ and $10.0\text{torr}$ for $P_1$ and $1.0\text{torr}$ for $P_2$ in the equation (3).

$\begin{array}{c}\Delta H_{\text{vap}}=-\frac{\left(8.314\text{ J/K}\cdot \text{mol}\right)\ln \left(\frac{1.0\text{torr}}{10.0\text{torr}}\right)}{\left(\frac{1}{327.5\text{ K}}-\frac{1}{368.5\text{ K}}\right)}\\ =56350.28\text{ J/mol}\\ \approx 5.6\times {10}^4\text{J/mol}\end{array}$

Substitute $25°\text{C}$ in equation (2) to calculate the temperature $T_3$ in Kelvin as follows:

$\begin{array}{c}{T}_3\left(\text{K}\right)=25°\text{C}+273\\ =298\text{ K}\end{array}$

The expression to calculate $P_3$ is as follows:

$\ln \left(\frac{P_3}{P_1}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_3}-\frac{1}{T_1}\right)$        (4)

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $368.5\text{ K}$ for $T_1$, $298\text{ K}$ for $T_3$ and $56350.28\text{ J/mol}$ for $\Delta H_{\text{vap}}$ in the equation (4).

$\begin{array}{c}\ln \left(\frac{P_3}{P_1}\right)=\frac{-\left(56350.28\text{ J/mol}\right)}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{298\text{ K}}-\frac{1}{368.5\text{ K}}\right)\\ =-4.351324\end{array}$

Take the exponential of $-4.351324$ and substitute $10.0\text{torr}$ for $P_1$ as follows:

$\begin{array}{c}\frac{P_3}{10.0\text{torr}}=e^{-4.351324}\\ \frac{P_3}{10.0\text{torr}}=0.0128897\\ P_3=0.128897\text{torr}\\ \approx 0.13\text{torr}\end{array}$

Conclusion

The vapor pressure of methyl salicylate at $25\text{°C}$ is $0.13\text{torr}$.

(b)

Interpretation Introduction

Interpretation:

The volume required at $25\text{°C}$ to vaporize $1.00\text{mg}$ of methyl salicylate is to be determined.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$        (5)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

Answer to Problem 12.129P

The volume required at $25\text{°C}$ to vaporize $1.00\text{mg}$ of methyl salicylate is $0.95\text{L}$.

Explanation of Solution

The expression to calculate the moles of methyl salicylate is as follows,

$\text{Moles of methyl salicylate}=\frac{\text{Given mass of methyl salicylate}}{\text{Molar mass of methyl salicylate}}$        (6)

Substitute $1\text{mg}$ for a given mass of methyl salicylate and $152.16\text{ g/mol}$ for molar mass of methyl salicylate in the equation (6).

$\begin{array}{c}\text{Moles of methyl salicylate}=\frac{1\text{mg}}{152.16\text{ g/mol}}\left(\frac{{10}^{-3}\text{g}}{1\text{mg}}\right)\\ =0.006572\times {10}^{-3}\text{mol}\end{array}$

Rearrange the equation (5) to calculate the volume of methyl salicylate.

$V=\frac{nRT}{P}$        (7)

Substitute the value $0.128897\text{torr}$ for $P$, $298\text{ K}$ for $T$, $0.006572\times {10}^{-3}\text{mol}$ for $n$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (7).

$\begin{array}{c}V=\frac{\left(0.006572\times {10}^{-3}\text{mol}\right)\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(298\text{ K}\right)}{0.128897\text{torr}}\left(\frac{760\text{torr}}{1\text{atm}}\right)\\ =0.94805\text{L}\\ \approx 0.95\text{L}\end{array}$

Conclusion

The volume required at $25\text{°C}$ to vaporize $1.00\text{mg}$ of methyl salicylate is $0.95\text{L}$.