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: | ||

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 = nRT putting the value of PV, we get, | ||

nRT= 1/3 (mN) | ||

Since number of mole ( n) = molecules/Avogadro’s numbernumber of mole ( n) = N/N_{A} | ||

Therefore, | ||

[N/N_{A}] R T= 1/3 (mN)N_{A} R T= 1/3 (m)3 [N_{A} R] T= (m) | ||

But N = Boltzman’s constant (K), thus,_{A} R | ||

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)_{av}therefore, | ||

(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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