Carbon tetrachloride has an enthalpy of vaporization of 29.82kj/mol. If CCl4 has a normal boiling point of 350k, what is its vapor pressure at 273k? (R=8.31 J/molK)
Use the Clausius-Clapeyron equation.
To find the vapor pressure of carbon tetrachloride (CCl4) at 273 K, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 is the vapor pressure at the boiling point (350 K)
P2 is the vapor pressure at the desired temperature (273 K)
ΔHvap is the enthalpy of vaporization (29.82 kJ/mol)
R is the gas constant (8.31 J/molK)
T1 and T2 are the boiling point and desired temperature, respectively, in Kelvin.
To use the equation, we need to convert the enthalpy of vaporization from kJ/mol to J/mol. Since 1 kJ = 1000 J, the enthalpy of vaporization becomes 29.82 * 1000 = 29820 J/mol.
Now we can plug the values into the equation:
ln(P2/760) = -(29820/8.31) * (1/273 - 1/350)
To solve for P2, we'll rearrange the equation:
P2/760 = e^(-(29820/8.31) * (1/273 - 1/350))
Now we can calculate P2.