boiling points, melting points and absorbed radiation Q's

koomanchoo

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/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.

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 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

Attached Files

Data values for 4.doc (23.0 KB, 4 views)