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How can you determine the quality of a vapor/liquid mixture if you know the specific enthalpy? Is there an equation for this somewhere?

If you only have the pressure and temperature data, but no information about enthalpy, could you find the quality? Is there an question somewhere?

2007-02-08 11:14:08 · 5 answers · asked by Ooze90 3 in Science & Mathematics Engineering

5 answers

There is a short answer to this. The enthalpy of a liquid/vapor mixture is just a linear combination of the enthalpy of the two components. If the specific enthalpy of the liquid phase is hf and the vapor phase is hg then the relation between the mixture specific enthalpy (h) and the quality (x) is just:

h = x * hg + (1-x) * hf

That's it. As long as you know all three enthalpies, you can solve for quality.

Your second question involves figuring quality from pressure and temperature. The answer is "yes and no". First, as usual, assume equilibrium conditions. In that case, if your pressure temperature conditions are below the saturation line the quality is 0. If the conditions are above the saturation line, the quality is 1. Unfortunately, we usually are only interested in quality only at the saturation point when both liquid and vapor can exist.

In that case, pressure and temperature alone cannot determine quality. One other parameter is needed, if could be specific volume, enthalpy or entropy as long as you know the corresponding values for the liquid and vapor separately. These are usually well tabulated. The equation I gave above works for any of these properties just as well as for enthalpy.

2007-02-08 18:08:30 · answer #1 · answered by Pretzels 5 · 2 0

Specific Enthalpy

2016-12-17 04:08:12 · answer #2 · answered by pilkington 4 · 0 0

Formula For Enthalpy

2016-10-02 12:12:56 · answer #3 · answered by ladwig 4 · 0 0

The first part was well answered previously.

For the second part, there it is not possible. The only way is to use an "expansion calorimeter" which allows to expand the vapor until open pressure. Whit the temperarture and assuming isentalpic expansion, it is possible to obtain the enthalpy, relating to quality.

2007-02-08 19:34:40 · answer #4 · answered by Alonso J 2 · 0 0

The standard enthalpy change of formation is measured in units of energy per amount of substance. Most are defined in kilojoules per mole, or kJ mol-1, but can also be measured in calories per mole, joules per mole or kilocalories per gram (any combination of these units conforming to the energy per mass or amount guideline). In physics the energy per particle is often given in electronvolt which corresponds to about 100 kJ mol-1.

All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation.

The standard enthalpy change of formation is used in thermochemistry to find the standard enthalpy change of reaction. This is done by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products, as shown in the equation below.

ΔHreactionO = ΣΔHfO (Products) - ΣΔHfO (Reactants)

The standard enthalpy of formation is equivalent to the sum of many separate processes included in the Born-Haber cycle of synthesis reactions. For example, to calculate the standard enthalpy of formation of sodium chloride, we use the following reaction:

Na(s) + (1/2)Cl2(g) → NaCl(s)
This process is made of many separate sub-processes, each with their own enthalpies. Therefore, we must take into account:


Standard enthalpy change of formation Born-Haber diagram for lithium fluoride.The standard enthalpy of atomization of solid sodium
The first ionization energy of gaseous sodium
The standard enthalpy of atomization of chlorine gas
The electron affinity of chlorine atoms
The lattice enthalpy of sodium chloride
The sum of all these values will give the standard enthalpy of formation of sodium chloride.

Additionally, applying Hess's Law shows that the sum of the individual reactions corresponding to the enthalpy change of formation for each substance in the reaction is equal to the enthalpy change of the overall reaction, regardless of the number of steps or intermediate reactions involved. In the example above the standard enthalpy change of formation for sodium chloride is equal to the sum of the standard enthalpy change of formation for each of the steps involved in the process. This is especially useful for very long reactions with many intermediate steps and compounds.

Chemists may use standard enthalpies of formation for a reaction that is hypothetical. For instance carbon and hydrogen will not directly react to form methane, yet the standard enthalpy of formation for methane is determined to be -74.8 kJ mol-1 from using other known standard enthalpies of reaction with Hess's law. That it is negative shows that the reaction, if it were to proceed, would be exothermic; that is, it is enthalpically more stable than hydrogen gas and carbon.

It is possible to predict heat of formations for simple unstrained organic compounds with the Heat of formation group additivity method.

2007-02-08 11:23:05 · answer #5 · answered by nra_man58 3 · 0 0

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