10 grams of ethanol are placed in a 10-L sealed container at 292.15 K. At this t

Question

10 grams of ethanol are placed in a 10-L sealed container at 292.15 K. At this temperature, the equilibrium vapor pressure in the container is 10 torr. If we assume that the enthalpy of vaporization of ethanol is 43 kJ/mol and constant, compute the equilibrium vapor pressure of ethanol inside the container as a function of temperature from 300 K up to the temperature at which all of the ethanol is in the vapor phase. Plot the pressure inside the container as a function of temperature over the range 300 K

Explanation / Answer

If all the ethanol is in the vapor phase, pressure = nRT/V= (10/46)*0.0821*293.15/10=0.52 atm =0.52*760 Torr=395.2 Torr

from Classuies equation , ln (P2/P1) = (detlaH/R)*(1/T1-1/T2)

the temperature corresponding to 0.52 atm, ln (395.2/10)= (43*1000/8.314)*(1/293-1/T2), T2= 370K

So temperature between 300 to 370 need to be choosen

where T1= 293.15K and deltaH= 43 Kj/mol and R= 8.314 joules/mole.K

T2 is the temperature in K and P2= vapor pressure corresponding to the temperatuere T2

assme some temperature and calculate the P2 at that temperature. The results are tabulated.

Pressur varies from 10 Torr to 395.2 Torr

Assem some pressureP2. Calculate the corresponding temperature.

ln (P2/10)=   (43*1000/8.314)*(1/293.15-1/T2)

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