Question

The normal boiling point of a liquid is 282 °C. At what temperature (in

Answer 1

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

$ln\left (\frac{p_{2}}{p_{1}} \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}} \right )$

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

$\Delta H_{vap}$ = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

$ln\left (\frac{0.2}{1} \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15} \right )$

$\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}$

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.