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The distribution function for speeds of particles in an ideal gas can be expressed as f(v)= 87 m 2akg T |3/2 m e==mv . Prove that f(v) is maximum for €=kgT where From this results, show that the most probable speed is |2kgT m

The distribution function for speeds of particles in an ideal gas can be expressed as f(v)= 87 m 2akg T |3/2 m e==mv . Prove that f(v) is maximum for €=kgT where From this results, show that the most probable speed is |2kgT m

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The distribution function for speeds of particles in an ideal gas can be expressed as |3/2 m -e/k,T ee f(v)=- m 2a kg T Prove that f(v) is maximum for e=kgT where E=-mv |2 kgT From this results, show that the most probable speed is v,=N m
Transcribed Image Text:The distribution function for speeds of particles in an ideal gas can be expressed as |3/2 m -e/k,T ee f(v)=- m 2a kg T Prove that f(v) is maximum for e=kgT where E=-mv |2 kgT From this results, show that the most probable speed is v,=N m
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a) to maximise the f(v)

                              f(v) =Cεe-εKBT                                    (C =8πm(m2πKBT))df(ν)dε =0apply the chain rule,0=e-εKBT + ε(-1kBT)e-εKBTafter solving this 'ε=kBT

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