The total pressure in a flask containing air and ethanol at 25.7C is 878 mm Hg. The pressure of the air in the flask at 25.7C is 762 mm Hg. If the flask is immersed in a water bath at 450C, the total pressure is 980 mm Hg. The vapor pressure of ethanol at the new temperature is ? mm Hg.
Hint: you will need to correct the pressure of air at the new temperature using the Gas Law: P1/T1 = P2/T2
note: am i right in thinking that the original vapor pressure is 120? please help me from there :)
Are you sure the 980 numbeer isn't 1980?Your numbers don't make sense. Heating the air that much would have to raise the total pressure more.
To find the vapor pressure of ethanol at the new temperature, we need to use the given information and the ideal gas law.
Let's start by correcting the pressure of the air in the flask at the new temperature using the ideal gas law:
P1 / T1 = P2 / T2
Where:
P1 = 762 mm Hg (initial pressure of air)
T1 = 25.7 °C (initial temperature)
P2 = unknown (pressure of air at the new temperature)
T2 = 45 °C (new temperature)
We can rearrange the equation to solve for P2:
P2 = (P1 * T2) / T1
P2 = (762 mm Hg * 45 °C) / 25.7 °C
P2 ≈ 1345 mm Hg
Now, let's find the vapor pressure of ethanol at the new temperature using Dalton's Law of Partial Pressures:
Total pressure = Pressure of air + Vapor pressure of ethanol
980 mm Hg = 1345 mm Hg (pressure of air) + Vapor pressure of ethanol
Vapor pressure of ethanol = 980 mm Hg - 1345 mm Hg
Vapor pressure of ethanol ≈ -365 mm Hg
It seems that the calculated vapor pressure of ethanol at the new temperature is -365 mm Hg, which doesn't make sense. Please double-check the given information or confirm if there may be any errors.