Chemistry Week 6

Chemistry Week 6 - Unit # 2 1 (3) Liquid-Vapor Equilibrium...

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Unit # 2 1 --- States of Matter --- (3) Liquid-Vapor Equilibrium (Vapor Pressure) If you leave a tray of water standing in your room (to humidify the room), eventually all of the water evaporates. Yet, the temperature of your room is nowhere close to the normal boiling point of water. This illustration hopes to substantiate the point that above the surface of a liquid, there is some vapor phase present. If the liquid is placed in an isolated container at some temperature with empty space above the liquid and a pressure gauge to measure the pressure of the gas, some molecules in the liquid will break away and form a vapor above the liquid phase. Equilibrium between the liquid and the gas is achieved when the driving force for losing molecules from the gas into the liquid equals the driving force for losing molecules from the liquid into the gas . This “driving force” is called the chemical potential . At equilibrium, the measured pressure is called the vapor pressure of the liquid at the temperature of the system. It takes energy to convert a liquid into a gas: Liquid + Heat Gas This energy is called the heat of vaporization , symbolized by H vap , with units kJ/mol. The relationship between vapor pressure ( p ) and temperature ( T ) is shown in the phase diagram, e.g., the curve (highlighted in black) representing the equilibrium between LIQUID and GAS .
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Unit # 2 2 --- States of Matter --- This curve is “pressure is a function of temperature,” p ( T ), but what is this function? Thermodynamics relies on studying changes, and establishing relationships between the various mechanical variables ( p , V ) and energy variables ( T , heat, …). To obtain this relationship, we use a relationship between the slope of this curve at the point ( T , p ), and the heat and volume changes involved when going between the liquid and the gas: The vapor pressure of a gas above a liquid and the temperature (K) are related by the Clausius- Clapeyron equation. The thermodynamic relationship between the slope of the curve p ( T ) and the heat and volume changes taking place: molar molar slope of ( ) curve dp H pT dT T V == ; where H molar = molar heat of the process and V molar = change in molar volumes for the process. For the equilibrium between liquid and gas, the process is written as Liquid + Heat ZZX YZZ Gas, H molar = H vap and V molar = V m (gas) V m (liquid), where V m (state) is the molar volume (L/mol) for the state of the substance. Since 1 mole of gas occupies so much more volume than 1 mole of liquid, V molar = V m (gas) V m (liquid) ~ V m (gas). Using the ideal gas law for one mole of gas, V m = RT/p , and we have, therefore, SOLID LIQUID GAS PRESSURE TEMPERATURE SLOPE of p ( T ) curve ( T b , 1 atm) ( T , p )
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Unit # 2 3 --- States of Matter --- () vap vap 2 Hp H dp dT T RT p RT ∆∆ == or (rearranging) vap 2 H dp dT p RT = ( R = 8.314 J/mol K).
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Chemistry Week 6 - Unit # 2 1 (3) Liquid-Vapor Equilibrium...

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