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PRABHU JAGATBANDHU COLLEGE DEPARTMENT OF CHEMISTRY For B.Sc 2 nd year(General)-2014-2015 Presented by Dr. Sipra Roy.

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Presentation on theme: "PRABHU JAGATBANDHU COLLEGE DEPARTMENT OF CHEMISTRY For B.Sc 2 nd year(General)-2014-2015 Presented by Dr. Sipra Roy."— Presentation transcript:

1 PRABHU JAGATBANDHU COLLEGE DEPARTMENT OF CHEMISTRY For B.Sc 2 nd year(General)-2014-2015 Presented by Dr. Sipra Roy

2 KINETIC THEORY OF GAS

3 STATE OF MATTER GASEOUS STATE SOLIDLIQUID GAS IDEAL GAS REAL GAS

4 Kinetic Theory Of Gas  To explain the behaviour of gases Bernoueli, Boltzman, Maxwell, Clausius, Van der Waaals (1738) imagined a model known as Kinetic Theory of Gases. Postulates of the Kinetic theory  Every gas consists of tiny discrete particles, called molecules  Gas molecules are always in random motion  During motion they collide with each other and also with the wall of the vessel.

5  Collision is perfectly elastic  Pressure of a gas is due to the bombardment of the molecules on the wall of the vessel  There is no force of attraction or repulsion among the gas molecules  Gas molecules are point mass  K.E of a gas is directly proportional to absolute temperature

6 The pressure of a gas x z y c l l l n = total no of gas molecules m = mass of a single molecule c = r.m.s. velocity u v w c 2 = u 2 + v 2 + w 2 Along x axis, change of momentum / collision = mu-(-mu) = 2mu no of collision /sec = u/l rate of change of momentum/molecule = 2mu 2 /l

7 Similarly, along y axis, momentum change /molecule/sec = 2mv 2 /l along z axis, momentum change/molecule/sec = 2mw 2 /l {In any arbitary direction momentum change/sec/molecule =2m{u 2 +v 2 +w 2 }/l = 2mc 2 /l Total momentum change/sec or Force = 2mnc 2 /l Total area of 6 walls = 6l 2 Pressure P = F/A = 2mnc 2 /l.6l 2 = mnc 2 /l 3 or PV = 1 3 mnc 2 l 3 =vol. of the cube = V3 ρ = mass density 1 1 or P = ρc 2 3

8 Gas Laws from Kinetic Theory Boyle’s Law & Charle’s Law: PV = or PV = n mnc 2 1 3 2 1 3 2 mc 2 3 2 or PV = (total K. E) But K.E α T So if T is constant, PV = constant This is Boyle’s Law When P is constant, then V α T This is Charle’s Law

9 Avogadro’s Law: Two different gases at the same pressure P, volume V, and same temperature T Let no of molecules are n 1 and n 2 respectively and mass of each molecules are m 1 and m 2 respectively. For first gas PV = For second gas PV = So, m 1 n 1 c 1 2 = m 2 n 2 c 2 2 ………………………………..(i) Since, T is same, so K.E is same So,, m 1 c 1 2 = m 2 c 2 2 ………………………………..(ii) Comparing (i) & (ii) n 1 = n 2 This is Avogadro,s Law m1n1c12m1n1c12 1 3 m2n2c22m2n2c22 1 3

10 Graham’s Law: 1 3 ρc 2 P = or c = √ 3P ρ or c = 1 √ρ [at constant P] Again, mnc 2 1 3 PV = For 1 mole gas Mc 2 1 3 PV = [n = N o and m.N o = mol. Wt = M] or c = 1 √M√M [at constant P & V] This is Graham’s Law

11 Expression for K.E : For 1 mole gas, mN o c 2 1 3 PV = = RT 2 1 3 2 mc 2 NoNo or = RT or 3 2 N o (K.E /molecule) = RT or 3 2 (K.E /mole) = RT or(K.E /mole)= 3 RT 2

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