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 45.0C, 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
This is the third time I have seen this question posted here. I see you changed the water bath temperature to something that makes sense: 45 C instead of 450. I explained how to to do this several days ago. The method remains the same.
The partial pressure air increases by a factor (273+45)/(273 + 25.7)= 1.0646, making the new partial pressure of air
762*1.0646 = 811 mmHG. The new partial pressure of alcohol is then 980-811 = 169 mmHg. This is higher than it was at the lower temperature, which makes sense for a change.
To find the vapor pressure of ethanol at the new temperature, we can use the ideal gas law equation: P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, P2 is the final pressure, and T2 is the final temperature.
Given:
Initial total pressure (P1) = 878 mm Hg
Initial air pressure (P1_air) = 762 mm Hg
Initial temperature (T1) = 25.7°C (which is 298.85 K)
Final total pressure (P2) = 980 mm Hg
Final temperature (T2) = 45.0°C (which is 318.15 K)
Step 1: Calculate the corrected pressure of air at the new temperature.
Using the ideal gas law equation, we can equate the initial and final pressures of air:
P1_air/T1 = P2_air/T2
P2_air = (P1_air * T2) / T1
P2_air = (762 mm Hg * 318.15 K) / 298.85 K
P2_air = 813.45 mm Hg
Step 2: Calculate the vapor pressure of ethanol at the new temperature.
Since the total pressure at the new temperature is given as 980 mm Hg and the pressure of air at the new temperature is 813.45 mm Hg, we can subtract the pressure of air from the total pressure to find the vapor pressure of ethanol:
Vapor pressure of ethanol = P2 - P2_air
Vapor pressure of ethanol = 980 mm Hg - 813.45 mm Hg
Vapor pressure of ethanol = 166.55 mm Hg
Therefore, the vapor pressure of ethanol at the new temperature is 166.55 mm Hg.