The total pressure in a flask containing air and ethanol at 257C is 878 mm Hg. The pressure of the air in the flask at 257C is 762 mm Hg. If the flask is immersed in a water bath at 300C , 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: P1T1 = P2T2
This question as been asked, and answered, mostly by DrWLS, at least a dozen times in the last week. All of the questions are variations of the same. He has always pointed out that it isn't likely to have a water bath at 300 C although most of the posts have been at 450 C. How do you get water to boil at a temperature higher than 100 C in a water bath?
the question actually asks at 25.7C and 30.0C
To find the vapor pressure of ethanol at the new temperature, we need to correct the pressure of air at the new temperature using the given equation.
First, let's identify the given values:
P1 = 762 mm Hg (pressure of air in the flask at 25°C)
T1 = 25°C (initial temperature)
P2 = 980 mm Hg (total pressure in the flask when immersed in a water bath at 30°C)
T2 = 30°C (new temperature)
Now, we can use the gas law equation P1T1 = P2T2 to solve for P2:
P1T1 = P2T2
Substituting the given values:
762 mm Hg * 25°C = 980 mm Hg * 30°C
To find the value of P2, we rearrange the equation:
P2 = (P1T1) / T2
P2 = (762 mm Hg * 25°C) / 30°C
P2 = 762 mm Hg * (25/30)
P2 = 762 mm Hg * 0.833
P2 = 635.646 mm Hg
So, the corrected pressure of air in the flask at 30°C is approximately 635.646 mm Hg.
Since the total pressure in the flask at 30°C is given as 980 mm Hg, we can subtract the corrected pressure of air from the total pressure to find the vapor pressure of ethanol at the new temperature:
Vapor pressure of ethanol = Total pressure - Pressure of air
Vapor pressure of ethanol = 980 mm Hg - 635.646 mm Hg
Vapor pressure of ethanol = 344.354 mm Hg
Therefore, the vapor pressure of ethanol at the new temperature (30°C) is approximately 344.354 mm Hg.