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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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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 = 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 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 number
number 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)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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