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;
[tex]ln\left (\frac{p_{2}}{p_{1}} \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}} \right )[/tex]
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
[tex]\Delta H_{vap}[/tex] = 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;
[tex]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 )[/tex]
[tex]\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}[/tex]
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.
