Temperature Effects On The Solubility Of Gases
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The solubility of gases is not a constant, unchanging condition. In diversified environments, where temperatures differ, the solubility of gases can also differ. Additionally, the solvent, or the greatest quantity of substance that is mixed with a gas to form a solution, can also effect a gas's solubility, or its ability to become dissolved and in turn contribute to a formed amount of concentration.
The Effects on the Solubility of Gases in the Universal Solvent
The solubility of gases is dependent on temperature. An increase in temperature results in a decrease in gas solubility in water, while a decrease in temperature results in an increase of gas solubility in water. To comprehend this phenomena one must consider the two processes that occur when a non-polar gas is added to water. Initially a type of cavity develops when adding the solute to the solvent, representative of the conformation and overall size of the added gas, and in turn a successive process occurs in which attractive forces between the gas and water molecules are stimulated. It is this dual process that induces the water to produce both attractive and repulsive forces. By examining the water on a microscopic level and the components of the water that portray propinquity to the non-polar gases, temperature dependencies become observable.
The Survival of Fish
The solubility of gases in water plays a major role in the survival of ectotherms, such as fish. Since an increase in temperature contributes to a decline in the solubility of gases in water, water that is of a higher temperature lacks oxygen. Therefore many fish are only capable of existing in waters of a lower temperature where more oxygen exists.
Barotrauma, commonly referred to as "the bends," is tissue damage that results from effects collocating with pressure, and thus the solubility of gases since pressure and gas solubility are directly proportional on the basis of Henry's Law, C=KP. Scuba divers are highly susceptible to the bends in that they subject their physiological pressure levels to those of the drastically diverse pressure levels of the ocean's waters. When under water, the existence of a vast amount of pressure causes the solubility of the gases in a diver's oxygen tank to incline; if a diver rises to the surface of the ocean too rapidly, where the pressure of the atmosphere is abated in comparison to the pressure within the water, and the solubility of the gases in the atmosphere is declined, injury of drastic measures can affect a diver.
Making Connections: The Phenomena of Gas Solubility
The reasoning for this relationship between temperature and gas solubility is similar to that of temperature and vapor pressure. An increase in temperature causes an increase in kinetic energy, resulting in a more rapid motion of molecules, breaking intermolecular bonds which enable molecules to escape from the solution. This is an example from the Second Law of Thermodynamics.
Scientific Evidence of the Effect of Temperature On Gas Solubility
When performing a study to interpret the effects of temperature control on the formation of soluble fermentation substances from sludge, it was found, after the performance of a series of experiments, that VFA generation through fermentation- increased by five-fold with a temperature change of 10-24 degrees Celsius (Cokgor,EU. Oktay,S. 2009).
Solubility: Link to Life
The solubility of gases also collocates with our life directly. At the University of Lund in Sweden, experimentation with infant formula led to the discovery that performing heat treatment on commercial formula affected the solubility and and ability to digest the milk proteins contained in the source of food (Boehm, G, Rhaiha,NC. 1994).
The Solubility of Gases in Organic Solvents
Gases assimilated with organic solvents become more soluble at higher temperatures. Le Chatelier's Principle can be applied to the determination of why solubility of gases increase with inclining temperatures. The principle proclaims that when a system is placed under stress, an equilibrium shift will occur in the direction that will most relieve the stress. In relation to the principle, adding heat to a solution will induce a shift in the equilibrium that favors dissolution in order to reduce heat. On a molecular stance, because organic solvents are incapable of forming hydrogen bonds with gases, in contrast to water, more heat is released when a gas is placed in water than in an organic solvent. A synthesized conclusion is that stronger attractions between a solvent and solute and entropy contribute to an greater transfer of heat, or enthalpy. Thus, increasing the temperature of the solution will favor an endothermic shift in the equilibrium of the system. And as a result the solubility off gases increases with increasing temperature in organic solvents.
The Exception to Gas Solubility
Noble gases, however, have a much more complex solubility. The solubility decreases with increased temperature, where it reaches a minimum at that certain temperature; then the solubility reverses directions, becoming more soluble at a higher temperature. A graph demonstrating the solubility of noble gases can be found here: http://www.sahra.arizona.edu/programs/isotopes/images/diagram6.gif
(x = not available)
1. You are given two beakers; each contains approximately 30ml of water. And you must determine which beaker contains the water with the higher solubility of oxygen. What information, in relation to the water is needed to make such a determination?
2. A system of cyclopentane and oxygen gas are at equilibrium with an enthalpy of -1234 KJ, predict whether the solubility of oxygen gas will be greater when heat is added to the system or when a temperature decrease occurs.
3. On the periodic table, which family, or group of elements displays complex solubility behavior? Clicking on the hyperlink under the section The Exception to Gas Solubility, determine which noble gas displays the greatest solubility at 50. C.
4. You purchase a rare species of plant that acquires oxygen through the solution it is placed in. Knowing that the plant will be kept outside over the scorching summer, what type of solution should the plant be placed in to assure that it receives the greatest amount of oxygen possible?
5. Determine the solubility of N2(g) when combined with H2O at .0345 atm. At 0 .C the pressure of N2 is 1.00 atm and its solubility is 21.0 ml at STP.
How did you do?
1. Being informed of the temperature of the solution of H2O within the beaker will allow for one to make that conclusion that the beaker possessing the higher temperature will consist of a gas of a declined solubility as a result of the intermolecular properties that cause gases to decline in solubilty with inclined temperatures when combined with water.
2. One should proximately determine that cyclopentane is an organic solvent. Oxygen gas and cyclopentane in a system at equilibrium, where the entropy is negative, will be be displaced from equilibrium when any type of temperature change is inflicted on the system. But because it is an incline in temperature that favors an endothermic reaction and the dissolution of the gas to reduce the addition of heat to the system, it is the incorporation of more heat into the system that will augment the solubility of the gas.
3. noble gases; Helium ( with a solubility of 1).
4. The plant should be placed in a solution that consists of an organic solvent to assure that, with the augmentation of the outside temperature, the solubility of oxygen in the solution will also increase.
5. 3.29*10-5M N2
Begin by determining the molarity ( solubility) of N2 (g) at O . C and STP.
NOTE: Molarity is equivalent to: mols of solute/ L of solution
At STP 1mol=22L
Molarity of N2 = 21ml *( 1L/ 1000ml)=.021L N2
=.021L N2*(1mol/22L)=.000954mols/ 1L
= 9.5*10-4 M N2
Now that the molarity of N2 0. C has been attained, Henry's Law constant k can be evaluated.
Henry's Law : C = k Pgas, where C is solubility, K is the constant, and Pis the partial pressure of the gas being considered.
Rearranging the formula to solve for k : k=C/Pgas
k= 9.5*10-4 M N2/ 1atm
Now substitute k and the partial pressure of N2 into Henry's law:
C= (9.5*10-4M N2)*(.o345)= 3.29*10-5 M N2
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