Question

Use the Clausius-Clapeyron equation to calculate the vapor pressure for the following samples. Assume the amount of each sample is 1 mole.
a. Water at 63 C.
b. Ethanol at 23 C.
c. Chloroform at 52 °C.

Answer 1

The vapor pressure of any liquid at its boiling point is 1 atm.

(a) for water, $P_1=1 atm,T_1=373 \ K,T_2=63\degree C=63+273=333\ K$

$\triangle H_{vap} =40.65KJ/mol$

By using Clausius-Clayperon equation,

$ln\frac{P_2}{P_1}=\frac{\triangle H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

$ln\frac{P_2}{1}=\frac{40.65\times 10^3}{8.314}(\frac{1}{373}-\frac{1}{333})$

$P_2=0.2071 \ atm$

(b) for ethanol, $P_1=1atm,T_1(B.P.)=78.37 \degree C=351.37 K,$ $T_2=23\degree C=296 K$

$\triangle H_{vap}=38.6 KJ/mol$

By using Clausius-Clayperon equation,

$ln\frac{P_2}{P_1}=\frac{\triangle H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

$ln\frac{P_2}{1}=\frac{38.6\times 10^3}{8.314}(\frac{1}{351.37}-\frac{1}{296})$

$P_2=0.84441 \ atm$

(c) For chloroform,$P_1=1atm,T_1(B.P.)=61.2 \degree C=334.2 K,$ $T_2=52\degree C=325 K$

$\triangle H_{vap}=31.4 KJ/mol$

By using Clausius-Clayperon equation,

$ln\frac{P_2}{P_1}=\frac{\triangle H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

$ln\frac{P_2}{1}=\frac{31.4\times 10^3}{8.314}(\frac{1}{334.2}-\frac{1}{325})$

$P_2=0.72622 \ atm$