Another liquid has a vapour pressure of 20 kPa at 20oC and an enthalpy of vapourisation of 40 kJ mol-1.Calculate the normal boiling point (oC).
Question
Another liquid has a vapour pressure of 20 kPa at 20oC and an enthalpy of vapourisation of 40 kJ mol-1.Calculate the normal boiling point (oC).
Solution
To solve this problem, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at one temperature to the vapor pressure of the same substance at another temperature. The equation is as follows:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where: P1 = initial pressure = 20 kPa P2 = final pressure = 101.3 kPa (normal boiling point pressure) ΔHvap = enthalpy of vaporization = 40 kJ/mol = 40000 J/mol R = gas constant = 8.314 J/(mol*K) T1 = initial temperature = 20°C = 20 + 273.15 = 293.15 K T2 = final temperature = ? (this is what we're solving for)
Plugging in the known values, we get:
ln(101.3/20) = -40000/8.314 * (1/T2 - 1/293.15)
Solving for T2, we get:
T2 = 1 / ((ln(101.3/20) * -8.314 / 40000) + 1/293.15)
T2 ≈ 353.15 K
Converting this back to Celsius, we subtract 273.15:
353.15 K - 273.15 = 80°C
So, the normal boiling point of the liquid is approximately 80°C.
Similar Questions
The vapour pressure of a certain liquid is 30 kPa at 25oC and 60 kPa at 38oC. What is the enthalpy of vaporisation? (kJ mol-1)
A certain substance X has a normal boiling point of 146.3°C and a molal boiling point elevation constant =Kb·1.47°C·kgmol−1. Calculate the boiling point of a solution made of 6.083g of urea NH22CO dissolved in 200.g of X.Round your answer to 4 significant digits.
The normal boiling point of a liquid isQuestion 6Select one:the temperature at which the gas molecules have more kinetic energy than the molecules in the liquid.the temperature above which the substance cannot exist as a liquid regardless of the pressure.the temperature at which the liquid will usually boil.the temperature at which the vapour pressure equals 101.3 kPathe only temperature at which there can be equilibrium between liquid and gas.
The boiling point of CCl4 is 76.8°C. The heat of vaporization of CCl4 is 271 kJ/mol. How much heat is required to convert 25.0 g of liquid CCl4at 76.8°C to gas phase CCl4? The molar mass of CCl4 is 153.8 g/mol. Group of answer choices34 kJ44 kJ54 kJ24 kJ
A certain substance X has a normal boiling point of 101.6°C and a molal boiling point elevation constant =Kb·2.26°C·kgmol−1. A solution is prepared by dissolving some urea NH22CO in 350.g of X. This solution boils at 105.3°C. Calculate the mass of urea that was dissolved.Round your answer to 2 significant digits.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.