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
OK, looks like you corrected the 450 you had last time to 45.0. Use the method they describe to get the new partial pressure of air,
P2 = (T2/T1)*P1. Then subtract P2 from the total pressure of 980 mm Hg to get the vapor pressure of ethanol, since the rest of the gas present must be ethanol vapor.
We will be glad to critique your work
for P2, i got 1334.2. that seems wrong b/c its bigger than 980, the total pressure.
P ethanol at 25.7C is 878-762 = 116 mm Hg.
P1/T1 = P2/T2
P1 is pressure of ethanol at 25.7C, P2 is pressure of ethanol at 45C.
You have to convert the Celsius into Kelvin by adding 273, so T1 = 298.7 K (25.7 + 273) and T2 = 318 K.
then substitute the values to find P2 (pressure of ethanol at the new temperature)
To find the new partial pressure of air (P2), we can use the gas law equation P1/T1 = P2/T2.
Given:
P1 = 762 mm Hg (pressure of air at 25.7°C)
T1 = 298.7 K (25.7°C + 273)
T2 = 318 K (45.0°C + 273)
Plugging in the values:
P1/T1 = P2/T2
762 mm Hg / 298.7 K = P2 / 318 K
To solve for P2, we cross-multiply:
P2 = (762 mm Hg * 318 K) / 298.7 K
P2 ≈ 812.8 mm Hg
Now let's calculate the vapor pressure of ethanol at the new temperature by subtracting the partial pressure of air (P2) from the total pressure (980 mm Hg).
Vapor pressure of ethanol at the new temperature = Total pressure - P2
Vapor pressure of ethanol = 980 mm Hg - 812.8 mm Hg
Vapor pressure of ethanol ≈ 167.2 mm Hg
Therefore, the vapor pressure of ethanol at the new temperature is approximately 167.2 mm Hg.