10. toluene has a standard enthalpy of vaporization ($\\delta h_{vap}^0$) of 38.2 kj/mol and a standard…

Explanation:

Step1: Convert temperature to Kelvin

$T = 25.0 + 273.15=298.15\ K$

Step2: Calculate $\Delta G_{vap}^0$ at the normal - boiling - point

At the normal - boiling - point ($T_b$), $\Delta G_{vap}^0 = 0$. We know that $\Delta G_{vap}^0=\Delta H_{vap}^0 - T\Delta S_{vap}^0$. First, convert $\Delta H_{vap}^0$ to J/mol: $\Delta H_{vap}^0 = 38.2\times10^{3}\ J/mol$.

Step3: Use the Clausius - Clapeyron equation in the form $\ln\left(\frac{P_2}{P_1}\right)=-\frac{\Delta H_{vap}^0}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

At the normal - boiling - point, $P_1 = 1\ bar$ and $T_1$ is the temperature at which $\Delta G_{vap}^0 = 0$. Since $\Delta G_{vap}^0=\Delta H_{vap}^0 - T\Delta S_{vap}^0 = 0$, then $T_1=\frac{\Delta H_{vap}^0}{\Delta S_{vap}^0}=\frac{38.2\times 10^{3}\ J/mol}{100.3\ J\ K^{-1}\ mol^{-1}}\approx381\ K$. We want to find $P_2$ at $T_2 = 298.15\ K$, and $R = 8.314\ J\ K^{-1}\ mol^{-1}$. $\ln\left(\frac{P_2}{1}\right)=-\frac{38.2\times 10^{3}\ J/mol}{8.314\ J\ K^{-1}\ mol^{-1}}\left(\frac{1}{298.15\ K}-\frac{1}{381\ K}\right)$ First, calculate $\frac{1}{298.15\ K}-\frac{1}{381\ K}=\frac{381 - 298.15}{298.15\times381}=\frac{82.85}{113695.15}\approx7.29\times 10^{-4}\ K^{-1}$ Then, $-\frac{38.2\times 10^{3}\ J/mol}{8.314\ J\ K^{-1}\ mol^{-1}}\times7.29\times 10^{-4}\ K^{-1}\approx - 3.31$ So, $\ln(P_2)=-3.31$ $P_2 = e^{-3.31}\approx0.0352\ bar$

Answer:

a) 0.0352 bar