# SHOW THAT THE AVERAGE TRANSLATIONAL KINETIC ENERGY OF THE MOLECULES OF A GAS IS DIRECTLY PROPORTIONAL TO ABSOLUTE TEMPERATURE

 QUESTION Starting from the expression P = r 1/3  show that the Average translational kinetic energy of the molecules of a gas is directly proportional to absolute temperature. PROOF According to the kinetic equation of pressure of a gas: P = r 1/3 But r = density of gas r = density of gas = mass of gas / volume of gas r = density of gas = mN / VPutting the value of density (r) P = 1/3 (mN / V ) P V= 1/3 (mN) But PV = nRTputting the value of PV, we get, nRT= 1/3 (mN) Sincenumber of mole (n) = molecules/Avogadro’s numbernumber of mole (n) = N/NA Therefore, [N/NA] R T= 1/3 (mN) NA R T= 1/3 (m) 3 [NA R] T= (m) But NA R = Boltzman’s constant (K), thus, 3 K T= (m) Multiplying both sides by 1/2 (3/2) K T= (1/2) m (1/2) m = (3/2) K T But (1/2) m = average translational kinetic energy of gas molecules = (K.E)avtherefore, (K.E)av = (3/2) K T As (3/2) K is a constant term, therefore, (K.E)av = (constant) T

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