Question

1. (2 pts) the vapor pressure of ethanol is 115 torr at 34.9°c. if δh_vap of ethanol is 38.6 kj/mol, calculate the temperature (in °c) when the vapor pressure is 760 torr. r = 8.314 j/mol·k.

ln\frac{p_1}{p_2}=\frac{-δh_{vap}}{r}(\frac{1}{t_2}-\frac{1}{t_1})

Explanation:

Step1: Identify the values

Let $P_1 = 115$ torr, $T_1=(34.9 + 273.15)\text{K}=308.05\text{K}$, $P_2 = 760$ torr, $\Delta H_{vap}=38.6\times10^{3}\text{J/mol}$, $R = 8.314\text{J/mol}\cdot\text{K}$.

Step2: Rearrange the Clausius - Clapeyron equation

The Clausius - Clapeyron equation is $\ln\frac{P_1}{P_2}=-\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$. First, we can rewrite it to solve for $\frac{1}{T_2}$:
$\frac{1}{T_2}=\frac{1}{T_1}-\frac{R\ln\frac{P_1}{P_2}}{\Delta H_{vap}}$.

Step3: Substitute the values

$\ln\frac{P_1}{P_2}=\ln\frac{115}{760}=\ln(0.1513)= - 1.89$.
$\frac{R\ln\frac{P_1}{P_2}}{\Delta H_{vap}}=\frac{8.314\times(- 1.89)}{38.6\times10^{3}}=\frac{-15.713}{38.6\times10^{3}}=-4.07\times10^{-4}\text{K}^{-1}$.
$\frac{1}{T_1}=\frac{1}{308.05}\text{K}^{-1}\approx3.25\times10^{-3}\text{K}^{-1}$.
$\frac{1}{T_2}=3.25\times10^{-3}\text{K}^{-1}-(-4.07\times10^{-4}\text{K}^{-1})=(3.25\times10^{-3}+4.07\times10^{-4})\text{K}^{-1}=3.66\times10^{-3}\text{K}^{-1}$.

Step4: Solve for $T_2$

$T_2=\frac{1}{3.66\times10^{-3}\text{K}^{-1}}\approx273\text{K}$.

Step5: Convert to Celsius

$T_2$ in Celsius is $T = 273 - 273.15=- 0.15\approx0^{\circ}\text{C}$.

Answer:

$0^{\circ}\text{C}$