Show that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

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Ramakrishnan Chandrasekaran R

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Here is the proof for the question 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 / V Putting...
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Here is the proof for the question According to the kinetic equation of pressure of a gas: P =r1/3 Butr= density of gas r= density of gas = mass of gas / volume of gas r= density of gas = mN / V Putting the value ofdensity (r) P =1/3(mN / V) PV=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 number number of mole (n) =N/NA Therefore, [N/NA] R T=1/3(mN) NAR T=1/3(m) 3 [NAR] T=(m) ButNAR= 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)av therefore, (K.E)av= (3/2) K T As(3/2) Kis a constant term, therefore, (K.E)av= (constant) T read less
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