Answer : The vapor pressure of bromine at [tex]10.0^oC[/tex] is 0.1448 atm.
Explanation :
The Clausius- Clapeyron equation is :
[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = vapor pressure of bromine at [tex]10.0^oC[/tex] = ?
[tex]P_2[/tex] = vapor pressure of propane at normal boiling point = 1 atm
[tex]T_1[/tex] = temperature of propane = [tex]10.0^oC=273+10.0=283.0K[/tex]
[tex]T_2[/tex] = normal boiling point of bromine = [tex]58.8^oC=273+58.8=331.8K[/tex]
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 30.91 kJ/mole = 30910 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
[tex]\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})[/tex]
[tex]P_1=0.1448atm[/tex]
Hence, the vapor pressure of bromine at [tex]10.0^oC[/tex] is 0.1448 atm.
The vapor pressure of bromine at 10.0°C is 0.16 atm.
Using the formula;
ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)
P2 = ?
P1 = 1 atm
ΔHvap = 30.91 kJ/mol
T2 = 10.0°C + 273 = 283 K
T1 = 58.8°C + 273 = 331.8 K
Substituting values;
ln(P2/1) = -30.91 × 10^3/8.314 JK-1 mol-1 (1/283 K - 1/331.8 K)
ln(P2/1) = -1.859
P2 = e^-1.859
P2 = 0.16 atm
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