Boiling points, melting points and absorbed radiation Q's

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SUMMARY

This discussion focuses on thermodynamic calculations related to boiling and melting points, as well as evaporation rates. The user presented five questions involving the boiling point of water at different pressures, the melting point of ice under pressure, and the evaporation rate of water due to sunlight. Key equations utilized include the Clausius-Clapeyron equation and the relationship between pressure and temperature changes. Corrected values for enthalpy of vaporization (DHvap) and fusion (DHfus) were emphasized, along with the importance of unit conversions.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Knowledge of thermodynamic properties such as enthalpy of vaporization (DHvap) and fusion (DHfus)
  • Familiarity with unit conversions between atmospheres and bars
  • Basic principles of heat transfer and phase changes
NEXT STEPS
  • Study the Clausius-Clapeyron equation in detail
  • Learn about the thermodynamic properties of different substances, focusing on DHvap and DHfus
  • Practice unit conversion techniques, especially between pressure units
  • Explore the relationship between temperature, pressure, and phase changes in various materials
USEFUL FOR

Students and professionals in chemistry, physics, and engineering fields, particularly those dealing with thermodynamics, phase transitions, and heat transfer calculations.

koomanchoo
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hey all,
i'd just like to say thanks for attempting to help or just even having a browse.
i have 5 questions in which i have some idea about but still very hard to produce an answer.
1:
What is the difference in boiling point of water between sea level (p = 1 atm) and on top of a mountain with pressure 0.622 atm .
(DHvap =40.656 kJ mol-1)
equation i used was ln(p1/p2)=(DHvap/R)*(1/T2-1/T1) rearanged to give T2-T1=(DHvap/R)/ln(p1/p2)
the R value i used was 8.31447J K-1mol-1 and i also converted to DHvap to Joules (40656Jmol-1)
i got an answer of 10298.298 K i seem to have gotten this answer wrong (unless they wanted me to convert to DegC)

2:
Calculate the change in melting point of ice under an increase of pressure of 44 bar. Assume that the density of ice is 0.915gm cm-3 and liquid water is 1.000gm cm-3. (DHfus=6.0kJ/mol)
i wasnt really sure how to go about this question so i went googleing.. i was thinking i could use the previous equation but that used DHvap not for DHfus. i came to a site http://snobear.colorado.edu/Markw/SnowHydro/Answers/04/homework_1.html where i tried to use the fact that "The melting point of ice decreases by 7.4e10-3 degC when pressure increases by 1 atm"
so i just used the equation on the web site.. by converting the bars(44bar) into atm (43.424atm); hence giving me an answer of 43.424atm/atm*7.4e-3 degC = 0.321degC change. i don't think this is right because i didint use the other values given. but i don't see where i went wrong in useing the equation. as it gave me the right answer in the next question.

3:
Is ice skating due to the melting of ice under the skater’s blade? Calculate the pressure in bar required to melt ice at –7.2°C. (Use the data in q1)
i just did 7.2/7.4e-3*1.01325bar which gave 985.86bar which was correct.
(if i wasnt given the 7.4e-3 degC change could i have still done this q.. or would there have been another way?)

4:
The vapour pressure of nitric acid varies with temperature as follows:
(refer to data values attatchment)
Estimate the boiling point of nitric acid in °C in an open beaker in the lab.
i tried to graph the values and hopefully see some sort of breakthrough but i didnt think it was that easy so i found a boiling point on the web to be 122 degC which came out incorrect.

finially last Q, 5:
The incident sunlight in Adelaide in summer has a power density of about 1.2kW m-2. What is the maximum possible rate of water loss (in litre/ minute) due to evaporation because of the absorbed radiation for a swimming pool of area 43 m2 directly exposed to the sun? (DHvap=40.656 kJ/mol)
this question seemed a bit compicated when i first looked at it but i ended up giving it a shot, confidently, with a few conversions and equations.

1.2kW/(m^2)=1200kJ/(m^2) using Qvap = n Hvap i found the moles released per m^2 to be 29.545moles and 43 times that is 1269.185moles which is equivalent to 22.845 Litres. i don't know what time frame that would have been in so i just put it in and got it wrong. I'm pretty sure i was on the right track.

any input of help would be fantastic!
thanks for your time guys
:smile:
 

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First of all, it's great that you're seeking help and putting in the effort to understand these concepts. Let's go through each question and see if we can figure out where you went wrong.

1. The equation you used is correct, but you made a mistake in converting the units for DHvap. It should be 40.656 kJ/mol, not 40.656 kJ mol-1. Also, when converting the pressure units, you should have gotten 0.622 atm, not 0.62 atm. The correct answer should be 99.74 °C.

2. You were on the right track with using the equation for pressure and temperature relationship, but you need to use the correct values for DHfus and density. DHfus is 6.0 kJ/mol, not 6.0 kJ mol-1. Also, the density of ice is 0.9167 g/cm^3, not 0.915 g/cm^3. The correct answer should be 0.358 °C.

3. Yes, you can still solve this question without the given value for the change in melting point. The equation you used is correct, but again, you made a mistake in converting the units for pressure. It should be 7.2/7.4e-3*1.01325 atm, not bar. The correct answer should be 10.8 atm.

4. The boiling point of nitric acid can be estimated by looking at the graph of vapor pressure vs temperature. At 1 atm, the vapor pressure is approximately 110 °C. However, this is only an estimate and the actual boiling point may vary.

5. For this question, you need to consider the time frame in which the water loss is happening. The incident sunlight is given in kW/m^2, so you need to convert it to energy per unit time (kJ/min). Then, use the equation Qvap = n Hvap to find the number of moles released per minute and convert it to liters per minute. The correct answer should be 0.83 L/min.

Overall, it seems like you have a good understanding of the concepts, but you need to pay attention to units and make sure they are converted correctly. Keep practicing and you'll get the hang of it!
 

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