Answer:
The molar heat of vaporization of dichloromethane is 30.8kJ/mole
Explanation:
Using Clausius Clapeyron equation
ln (P1/P2) = (ΔHvap/R) (1/T2-1/T1)
At initial temperature of Ooc , the vapour pressure is 134mmHg
Therefore T1 = 0+273 =273K
And P1 = 134mmHg
At normal boiling point of 40oC , the vapour pressure is 760mmhg
T2 = 40 +273 = 313K
P2 = 760mmHg
ln (134/760) = ΔHvap/(8.3145 J/molK)
( 1/313K - 1/273K)
ΔHvap = 30800J/mol
= 30.8kJ/mol
Therefore, the molar heat of vaporization can be calculated using
Clausius Clapeyron equation using the above steps
Answer: The molar heat of vaporization is 30.81 kJ/mol
Explanation:
To calculate the molar heat of vaporization, we use the Clausius-Clayperon equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor used: 1 atm = 760 mmHg)
[tex]P_2[/tex] = final pressure = 134 mmHg
[tex]\Delta H_{vap}[/tex] = Molar heat of vaporization
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = [tex]40^oC=[40+273]K=313K[/tex]
[tex]T_2[/tex] = final temperature = [tex]0^oC=[0+273]=273K[/tex]
Putting values in above equation, we get:
[tex]\ln(\frac{134}{760})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{313}-\frac{1}{273}]\\\\\Delta H_{vap}=30814.6J/mol=30.81kJ/mol[/tex]
Hence, the molar heat of vaporization is 30.81 kJ/mol
