|May21-05, 07:37 PM||#1|
boiling points, melting points and absorbed radiation Q's
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.
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)
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/Sn...omework_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 dont think this is right because i didint use the other values given. but i dont see where i went wrong in useing the equation. as it gave me the right answer in the next question.
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?)
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 dont 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
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