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 38.8C, the
total pressure is 980 mm Hg. The vapor pressure of ethanol at the new temperature is mm Hg.
I got it P1/T1=P2/T2 will get you there
initial pressure of the air(P1) = 762mm Hg
P2 = ?
T1 = 25.7+273 = 298.7K
T2 = 38.8+273 = 311.8K
P2 =?
P2 = 762 mm Hg x 311.8K/298.7K = 795.42 mm Hg
so, pressure of ethanol = 980 - 795.42 = 184.58 mm Hg
Answer in three sig fig will be 184 mm Hg.
To find the vapor pressure of ethanol at the new temperature, we can use the Clausius-Clapeyron equation, which relates vapor pressure to temperature:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
P1 = initial pressure (762 mm Hg)
P2 = final pressure in the water bath (980 mm Hg)
T1 = initial temperature (25.7°C + 273.15 = 298.85 K)
T2 = final temperature (38.8°C + 273.15 = 312.95 K)
R = gas constant (0.0821 L * atm/mol * K)
ΔHvap = enthalpy of vaporization of ethanol
First, let's find the value of ΔHvap for ethanol. The enthalpy of vaporization can be found in reference books or online sources. In this case, let's assume the value of ΔHvap for ethanol is 38.56 kJ/mol.
Now, we can plug in the values into the equation and calculate the vapor pressure:
ln(980/762) = (38.56 * 10^3 J/mol / (0.0821 L * atm/mol * K)) * (1/298.85 K - 1/312.95 K)
Calculating this expression should give us the natural logarithm of the ratio of the final pressure to the initial pressure.
ln(980/762) = (464.018) * (-0.003941)
Now, we can solve for the natural logarithm value:
ln(980/762) = -1.8293
Finally, to find the vapor pressure, we need to rewrite the equation in terms of the vapor pressure of ethanol:
P2 = P1 * e ^ ((ΔHvap/R) * (1/T1 - 1/T2))
P2 = 762 mm Hg * e ^ (-1.8293)
Using the above expression and solving it will give us the vapor pressure of ethanol at the new temperature in the water bath.