|By Neogbc (Neogbc) on Monday, March 08, 2004 - 08:33 pm: Edit|
Consense 100.00g of Benzene gas, C6H6, at 80°C to liquid benzene at 25°C. Specific heat of liquid benzene is 1.72J/g°C, boiling point is 80°C, and heat of vaporization is 30.8kJ/mol.
How would I approach this problem???
|By Athlonmj (Athlonmj) on Monday, March 08, 2004 - 08:41 pm: Edit|
Ok, think about the problem first. First the gas needs heat to convert it to liquid form. Then it needs energy to change the temperatore from 80 degrees to 25 degrees.
You need to use m * c * delta t and m * Heat vaporization
(mass of benzene) * (heat of vaporization of benzene) + (mass of benzene) * (specific heat of benzene) * (change in temperatore [80-25])
That's the approach, the easy part is just crunching the numbers. Make sure you're doing it all in kJ or J and convert when necessary
|By Vsage3 (Vsage3) on Monday, March 08, 2004 - 08:45 pm: Edit|
Why would a gas ever need heat to conver to liquid form? Despite this minor discrepancy, the rest of what Athlonmj says is correct.
|By Vsage3 (Vsage3) on Monday, March 08, 2004 - 08:50 pm: Edit|
double post with a 5 minute lag time.
|By Vsage3 (Vsage3) on Monday, March 08, 2004 - 09:13 pm: Edit|
Neo, what exactly is the question. Find the change in heat required for that to happen?
|By Neogbc (Neogbc) on Monday, March 08, 2004 - 09:15 pm: Edit|
Find the amount of heat involved (delta H)
|By Vsage3 (Vsage3) on Monday, March 08, 2004 - 09:56 pm: Edit|
Ok yes do that athlon said. Make sense?
|By Athlonmj (Athlonmj) on Monday, March 08, 2004 - 09:59 pm: Edit|
Sorry I wasn't thinking. change heat to energy.
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