# Lec.5 Gaseous state. Let’s Review Gases at low pressures (gas particles are far apart) have following characteristics: V α 1/P (constant temperature,

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Lec.5 Gaseous state

Let’s Review Gases at low pressures (gas particles are far apart) have following characteristics: V α 1/P (constant temperature, fixed amount of gas) – Boyle’s Law (P1V1=P2V2)T,n V α T (constant pressure, fixed amount of gas) – Charles’ Law (V1/T1=V2/T2)P,n V α n (at constant temperature and pressure) – Avogadro’s Law. Could be written as V = constant. n (linear plot) (V1/n1 =V2/n2)P,T This means that if you double amount of gas molecules, volume would double if pressure is to remain constant.

Combining three gas laws given above, we obtain: P1V1/n1T1 = P2V2/n2T2 This is the Ideal Gas Law which approximates the behavior for all gases at atmospheric pressure and becomes increasingly accurate as pressure is decreased. Because PV/nT = constant value (R), the Ideal Gas Law, which explains behavior of ideal/perfect gas, is written as: PV = nRT

 For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone.  for example:  Suppose we take two one-liter flasks, the first filled with O 2 at P=150 mmHg and the second filled with N 2 at P =100 mmHg  If N 2 transferred to the one liter flask containing O 2 the total P of the mixture found to be 250 mmHg.  This observation led Dalton to discover this law:-

Definition:"The pressure due to mixture of gases in a definite volume equals the sum of the partial pressure of the of the gases present“ P total = P a +P b +P c +……… P a, P b, P c =partial pressure of gases a, b, c The partial pressure of a gas in a mixture of gases can be calculated from the total P, when number of moles is known. If total volume of V liters have n a.n b, n c moles of gases a, b, c. Total number of moles : n = n a +n b +n c ………..…… (1) According to Dalton's law : P=P a +P b +P c ……..………. (2) For each gas occupying separately total volume V : P a V= n a RT……………. (3) P b V=n b RT……………. (4) P c V=n c RT……………. (5)

By adding (3), (4), and (5) (p a +p b +P c ) V= (n a +n b +n c ) RT….. (6) Combining (6), (1), and (2) : PV=n R T………………….……...(7) Dividing (3) by (7): Pa / P = na / n P a = (n a / n) * P where :(n a / n) = mole fraction Then P b = (n b / n) *P P c = (n c / n) * P Partial Pressure = mole fraction * total Pressure

Solution: P O2 =112 atm P N2 = ? P H2 =101 atm P total =278 atm P total = P O2 + P N2 + P H2 278 atm = 112 atm + 101 atm + P nitrogen P nitrogen = 278 atm - (112 atm + 101 atm ) P nitrogen = 65 atm Example 1 : A mixture of oxygen, hydrogen and nitrogen gases exerts a total pressure of 278 atm. If the partial pressures of the oxygen and the hydrogen are 112 atm. and 101 atm. respectively, what would be the partial pressure exerted by the nitrogen.

GRAHAM'S LAW OF DIFFUSION: "At same temperature and pressure the rate of diffusion of gases are inversely proportional to the square roots of their densities". R 1 / R 2 = √ (d 2 / d 1 ) i.e. at same T, P, if we have two gases the light gas diffuse faster than heavy gas. For gases 1 & 2 R 1, R 2 : rate of diffusion of 1 & 2 d 1, d 2 : densities of 1 & 2 If divide density to density of H 2 D 1, D 2 : relative densities of 1 & 2 D= d / dH 2 R 1 / R 2 =√ (D 2 / D 1 ) M 1, M 2 : molecular weight of 1, 2 M = 2DR 1 / R 2 =√ (M 2 /M 1 )

Example (2): How many cm3 of H 2 will diffuse in the same time as 13 cm3 of O 2 diffuses Given: O= 16 H= 1 Solution RH 2 / RO 2 =√(MO 2 /MH 2 ) =√(32 / 2)=4 RH 2 / 13 = 4 RH 2 = 13*4=52cm 3

 The study of mass relationships in chemical reactions is called stoichiometry. To solve stoichiometry problems, you must first do two very important steps. 1) Write a balanced equation for the reaction. 2) Convert all amounts of products and/or reactants into moles.  The coefficients (or numbers) in front of each reactant and product in the balanced chemical reaction tells you the ratio of how much of each you will react/produce.  For example: the reaction of hydrogen gas (H2) with oxygen gas (O2) to form water (H2O). 2H2 + O2 ---> 2H2O

So, we need two molecules of H 2 to react with each one molecule of O 2, and also, that this will form 2 molecules of water. The ratio of H 2 to O 2 to H 2 O is 2 : 1 : 2 Example: If you burn 10 grams of methane, how many grams of CO2 & H 2 O will be produced? (C:12, O:16, H:1) In this reaction, methane gas (CH 4 ) gets burned in oxygen (O 2 ) to form carbon dioxide (CO 2 ) and water vapor (H 2 O). CH 4 + 2O 2 --> CO 2 + 2H 2 O 1mole 2moles Number of moles(n)=Wt/M.wt M.Wt of CH 4 = 12+4*1=16 grams Number of moles of CH 4 =10/16=0.625 moles.  For CO 2 1mole of CH 4 1mole of CO 2 0.625 moles ?????

Number of moles of CO 2 = 0.625 moles M.Wt of CO 2 = 12+16*2=44 grams Number of moles(n)=Wt/M.wt Weight of CO 2 = 0.625 *44=27.5 grams  For H 2 O 1mole of CH 4 2moles of H 2 O 0.625 moles ????? Number of moles of H 2 O = 1.25 moles M.Wt of H 2 O = 2*1+16 =18 grams Number of moles(n)=Wt/M.wt Weight of H 2 O = 1.25 *18=22.5 grams Important note: Notice that the ratio of moles of CH 4 to CO 2 is 1 to 1, but the ratio of the weights is totally different. Remember, STOICHIOMETRY ONLY WORKS ON MOLES!

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