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 425C, 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
yo yo yo I got
175.1637945 b/c :
p1/t1 = p2/t2
762/298.85 = p2/ 315.65
p2 = 804.8362055
subtract w/ 980.... get the positive value....
175.1637945
your welcome. hahaha UCI!!!!
To find the vapor pressure of ethanol at the new temperature, we need to correct the pressure of air at the new temperature first.
Given:
Pressure of air at 25°C (P1) = 762 mm Hg
Total pressure at 25°C (Ptotal1) = 878 mm Hg
Using the gas law equation P1T1 = P2T2, we can calculate the corrected pressure of air at the new temperature.
Let's substitute the values into the equation:
P1 = 762 mm Hg
T1 = 25°C = 298 K
P2 = ? (pressure of air at the new temperature)
T2 = 42°C = 315 K
762 mm Hg * 298 K = P2 * 315 K
Now, solving for P2:
P2 = (762 mm Hg * 298 K) / 315 K
P2 = 719.47 mm Hg
We have now corrected the pressure of air to 719.47 mm Hg at the new temperature.
Next, to determine the vapor pressure of ethanol at the new temperature, we subtract the corrected pressure of air from the total pressure at the new temperature.
Total pressure at 42°C (Ptotal2) = 980 mm Hg
Pressure of air at 42°C (P2) = 719.47 mm Hg
Vapor pressure of ethanol at the new temperature (Pethanol2) = Ptotal2 - P2
Pethanol2 = 980 mm Hg - 719.47 mm Hg
Pethanol2 = 260.53 mm Hg
Therefore, the vapor pressure of ethanol at the new temperature is approximately 260.53 mm Hg.