Chapter 12, Problem 12.137P

Interpretation Introduction

Interpretation:

The concentration of amphetamine when it is in contact with $20°\text{C}$ air 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)

The ideal gas equation can be expressed as follows,

$PV=nRT$        (3)

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.

The conversion factor to convert $1\text{L}$ to $1{\text{m}}^3$ is as follows:

$1\text{L}={10}^{-3}{\text{m}}^3$

Answer to Problem 12.137P

The concentration of amphetamine when it is in contact with $20°\text{C}$ air is $2.9{\text{g/m}}^3$.

Explanation of Solution

Substitute $201°\text{C}$ in equation (2) to calculate the temperature $T_1$ in Kelvin as follows:

$\begin{array}{c}{T}_1\left(\text{K}\right)=201°\text{C}+273\\ =474\text{ K}\end{array}$

Substitute $83°\text{C}$ in equation (2) to calculate the temperature $T_2$ in Kelvin as follows:

$\begin{array}{c}{T}_2\left(\text{K}\right)=83°\text{C}+273\\ =356\text{ K}\end{array}$

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

$\Delta H_{\text{vap}}=-\frac{\left(R\right)\ln \left(\frac{P_2}{P_1}\right)}{\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$        (4)

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $474\text{ K}$ for $T_1$, $356\text{ K}$ for $T_2$ and $760\text{torr}$ for $P_1$ and $13\text{torr}$ for $P_2$ in the equation (4).

$\begin{array}{c}\Delta H_{\text{vap}}=-\frac{\left(8.314\text{ J/K}\cdot \text{mol}\right)\ln \left(\frac{13\text{torr}}{760\text{torr}}\right)}{\left(\frac{1}{356\text{ K}}-\frac{1}{474\text{ K}}\right)}\\ =48,370.199\text{ J/mol}\end{array}$

Substitute $20°\text{C}$ in equation (2) to calculate the temperature $T_3$ in Kelvin as follows:

$\begin{array}{c}{T}_3\left(\text{K}\right)=20°\text{C}+273\\ =293\text{ K}\end{array}$

The expression to calculate $P_3$ is as follows:

$\ln \left(\frac{P_3}{P_2}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_3}-\frac{1}{T_2}\right)$        (5)

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $293\text{ K}$ for $T_3$, $356\text{ K}$ for $T_2$ and $48,370.199\text{ J/mol}$ for $\Delta H_{\text{vap}}$ in the equation (5).

$\begin{array}{c}\ln \left(\frac{P_3}{P_2}\right)=\frac{-\left(48,370.199\text{ J/mol}\right)}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{293\text{ K}}-\frac{1}{356\text{ K}}\right)\\ =-3.5139112\end{array}$

Take the exponential of $-3.5139112$ and substitute $13\text{torr}$ for $P_2$ as follows:

$\begin{array}{c}\frac{P_3}{13\text{torr}}=e^{-3.51391124}\\ \frac{P_3}{13\text{torr}}=0.0297802\\ P_3=0.3871426\text{torr}\left(\frac{1\text{}\text{atm}}{760\text{torr}}\right)\\ \approx 5.09398\times {10}^{-4}\text{atm}\end{array}$

Rearrange the equation (3) to calculate the number of moles of amphetamine.

$n=\frac{PV}{RT}$        (6)

Substitute the value $5.09398\times {10}^{-4}\text{atm}$ for $P$, $293\text{ K}$ for $T$, $1{\text{m}}^3$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (6).

$\begin{array}{c}n=\frac{\left(5.09398\times {10}^{-4}\text{atm}\right)\left(1{\text{m}}^3\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(293\text{ K}\right)}\left(\frac{1\text{L}}{{10}^{-3}{\text{m}}^3}\right)\\ =0.02117612\text{mol}\end{array}$

The formula to calculate the mass of amphetamine is as follows:

$\text{Mass of amphetamine}=\left(\text{moles of}\text{amphetamine}\right)\left(\text{molar mass of amphetamine}\right)$        (7)

Substitute $0.02117612\text{mol}$ for moles of amphetamine and $135.20\text{g/mol}$ for molar mass of amphetamine in the equation (7).

$\begin{array}{c}\text{Mass of amphetamine}=\left(0.02117612\text{mol}\right)\left(135.20\text{g/mol}\right)\\ =2.8630114g\\ \approx 2.9\text{g}\end{array}$

Therefore, the mass of amphetamine in $1{\text{m}}^{\text{3}}$ is $2.9{\text{g/m}}^3$.

Conclusion

The concentration of amphetamine when it is in contact with $20°\text{C}$ air is $2.9{\text{g/m}}^3$.