20 grams of a pure liquid are in a closed 15 liter container at a temperature of

Question

20 grams of a pure liquid are in a closed 15 liter container at a temperature of 320 K. The molar heat of vaporization of the liquid is Delat H_vap = 12.500 kJ mol^-1. The molar mass of the liquid is 52.1 g mol^-1 and there are 0.086 moles of vapor in the container at equilibrium. Therefore, we expect that the equilibrium vapor pressure in the container will lie in the range 0.01 atm to 0.08 atm. lie in the range 0.08 atm to 0.16 atm. lie in the range 0.16 atm to 0.28 aim. lie in the range 0.28 atm to 0.49 atm. be greater than 0.49 atm.

Explanation / Answer

Number of moles of the solvent = Mass of pure liquid / molar mass

                                                = 20 g / 52.1 g mol-1

                                                = 0.384 moles

Given, mole fraction of vapor at equilibrium = 0.086

Total moles of vapor = mole fraction of vapor x total moles

                                    = 0.086 x 0.384 mol = 0.033 moles

Now, from the data available, the best way to proceed appears to be use of ideal gas equation.

Using ideal gas equation Ideal gas Law:        pV = nRT        - equation 1

            Where, p = pressure in atm = ?

            V = volume in L                      = 15.0 L

            n = number of moles = 0.033 mol    

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature      = 320 K                      

Or,

            P x 15.0 L = (0.033 mol) x (0.0821 atm L mol-1K-1) x (320 K)

            Or, P = 0.057768 atm.

Therefore, the vapor pressure of the vessel = = 0.057768 atm.

Correct option - A      (the value lies in range)

Note: Since, the temperature of the system is fixed and equilibrium is established, there is always 0.033 moles of vapor in the vessel (though a dynamic equilibrium). So, the net pressure exerted on vessel will be due to 0.033 mole of vapor (assumed to be ideal gas). Note: The Clausius-Clapeyron Equation of state is not required because T is fixed. The role of dH(vaporization) is to indicate (though straight forward) that 0.033 moles of the pure solvent (liquid phase) has already vaporized into vapor phase.    

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