Derive an expression for average kinetic energy of a molecule of an ideal gas.

Solution

Consider one mole of an ideal gas of molecular weight M_o at absolute temperature T.
Let,
m : mass of one molecule of gas
v : r.m.s speed of the gas molecule
P : Pressure exerted by an ideal gas

Therefore,

P = \frac{1}{3} {ρv}^{2} = \frac{1}{3} \frac{M}{V} v^{2}

That is,

PV =

\frac{1}{3} M v^{2}

... (1)

Also, PV = nRT
∴

\frac{1}{3} M v^{2} = n R T

\Rightarrow \frac{1}{2} \frac{M}{n} v^{2} = \frac{3}{2} RT \Rightarrow \frac{1}{2} M_{o} v^{2} = \frac{3}{2} RT \Rightarrow \frac{1}{2} {mNv}^{2} = \frac{3}{2} RT \Rightarrow \frac{1}{2} {mv}^{2} = \frac{3}{2} \frac{R}{N} T = \frac{3}{2} kT

Thus, the average kinetic energy of a molecule of an ideal gas is

\frac{3}{2} k T

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Mechanical Properties Of Fluids

Derive an expression for average kinetic energy of a molecule of an ideal gas.

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