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** Introduction**:

The ideal or perfect gas is illustrated in the postulates of the kinetic theory. The perfect or ideal gas doesn't exist. This is only hypothetical.

Real gases deviate from the ideal behavior and don't obey the gas laws perfectly. Assume a real gas subjected to low temperatures and very high pressures. The gas molecules are compressed to a very small volume and are not very energetic. Intermolecular forces of attraction example: Vander waals and dipole-dipole attractions come into play among the gas molecules. The volume of gas molecules will as well become important whenever compared by the space occupied through the gas.

** Dalton's Law**:

Dalton's law of partial pressures:

Dalton's law states that the total pressure applied via a mixture of gases in a container is equal to the sum of the partial pressures of the gases present in the mixture provided no chemical reaction takes place.

The partial pressure of a gas in a mixture is the pressure the gas would apply if it occupies the net volume of the mixture alone. Mathematically, for a mixture of three gases A, B, C in a container.

P_{Total} = P_{A} + P_{B} + P_{C}, here P_{Total }is the total pressure of the mixture and P_{A}, P_{B}, P_{c} are the partial pressures of A, B and C correspondingly.

Remember the general gas equation PV = nRT

For gas A, P_{A}V = n_{A}RT

For gas B, P_{B}V = n_{B}RT

For gas C, P_{C}V = n_{C}RT where n_{A}, n_{B} and n_{C} are the number of moles of A, B and C in the mixture.

P_{Total}V = (n_{A} + n_{B} + n_{C}) RT and

PA = x_{A} P_{Total}, P_{B} = x_{B }P_{Total} and P_{C} = x_{C }P_{Total}, where x_{A}, x_{B} and x_{C} are the mole fractions of A, B, and C in the mixture. This follows from the above that the partial pressure of a gas in a mixture is equivalent to the product of net pressure and the mole fraction of that gas in the mixture.

Dalton's law determines application in computations and determination comprising volumes of gases which contain water vapor as the case in the laboratory preparations of gases which are collected over water example: H_{2}, O_{2} and N_{2}.

Illustration:

A 1.0 dm^{3} gas jar consists of 0.02 mole oxygen and 0.125 mole nitrogen at 27^{o}C compute,

i) The partial pressures of oxygen and nitrogen.

ii) The net pressure of gas mixture R = 0.082 litre atm K^{-1} mol^{-1}

Solution: The general gas equation

PV = nRT

P = nRT/V, T = 273 + 27 = 300K

i) For O_{2}; PO_{2} = [0.02(0.82) x 300]/1.0 = 0.492 atm

For N_{2}; PN_{2} = [0.125 (.082) x 300]/1 = 3.075 atm

ii) P_{Total }= PO_{2} + PN_{2} = 0.492 + 3.075 = 3.567 atm

How the Kinetic theory illustrated Dalton's Law:

Whenever two or more gases mix without a chemical reaction, each gas behaves independently of the other as there are no attractive forces among the molecules. Each gas will apply its own pressure on the container wall. The net pressure of the gas mixture will thus be the sum of the pressures of the gases in the mixture.

** Graham's Law of Diffusion**:

Diffusion is the natural phenomenon. It comprises the movement of particles via a medium from a region of high particle concentration to a region of the lower particles concentration. Diffusion takes place in all the states of matter. By virtue of the very high kinetic energy of gas particles, diffusion is fastest in gases.

Graham's law deduces the relationship between the rate of diffusion of a gas (R) to its vapor density and illustrates why the law is known as Graham's Law of diffusion. The law defines that at a constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of density.

Mathematically;

R α 1/ρ, here 'R' is the rate of diffusion and 'ρ' the gas density.

From the relationship above, the higher the density of the gas the slower it will diffuse. If the rates of diffusion for two gases 1 and 2 are compared or divided the equation becomes:

R_{1}/R_{2} = √(ρ_{2}/ρ_{1}), here R_{1} R_{2} are rates and ρ_{1} ρ_{2} are densities correspondingly

Graham's law can as well be defined in terms of the relative molar masses of the gases

R_{1}/R_{2} = √(M_{2}/M_{1})

The rate of diffusion is inversely associated to the time of diffusion that is, the higher the rate of diffusion, the shorter the time for a particular mass of gas to diffuse. It obeys from the above that,

R_{1}/R_{2} = √(t_{1}/t_{2})

Here, t_{1} t_{2} are the times for similar mass of gases 1, 2 to diffuse under similar conditions.

** Avogadro's Law and its Applications**:

This law defines that equivalent volumes of all gases at similar temperature and pressure contain similar number of molecules.

Avogadro's law gives information on all number of molecules in a specified volume of gas and allows for change over from statements relating volumes of gases to statements relating to the number of molecules or moles.

Illustration: H_{2}(g) + Cl_{2}(g) → 2 HCl

1 vol of H_{2}(g) reacts with 1 volume of Cl_{2}(g) to give 2 volumes of HCl (g). Applying the Avogadro's law this means that 1 mole of H_{2}(g) reacts with 1 mole of Cl_{2}(g) to offer 2 moles of HCl(g). The application of law as allowed or established of formulae for gases. At 0^{o}C and 760 mm Hg pressure, 22.4 dm^{3} of any gas contains around 6.02 x 10^{23 }molecules that is, 1 mole of gas molecules. O^{o}C and 760 mm Hg pressure are termed to as standard temperature and pressure (S.T.P) correspondingly and 22.4 dm^{3} is the molar volume of gas at S.T.P.

** Gay Lussac's Law of Combining Volumes**:

This law defines that when gases react, they do so in volume that bear a simple ratio to the other and to other products (if gaseous) provided temperature and pressure remain constant.

Applying Avogadro's law, Gay Lussac's law can as well be limited as follows:

If gases react they do so in small whole numbers of molecules of reactants to generate small whole numbers of the products provided temperature and pressure remain constant.

Example: N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)

Gay Lussac one volume three volumes two volumes

Avogadro one molecule three molecules two molecules

From above, a balanced equation which can be written for the reaction given there is no unidentified product is:

N_{2}(g) + 3H_{2}(g) → 2 NH_{3}(g)

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