Average velocity maxwell boltzmann distribution
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The Maxwell-Boltzmann equation, which forms the basis of the kinetic theory of gases, defines the distribution of speeds for a gas at a certain temperature. From this distributionfunction, the most probable speed, the average speed, and the root-mean-square speed can be derived
If one has a probability density function P(x)P(x), then the expectation value of a quantity f(x)f(x) is given by
⟨f⟩=∫f(x)P(x)dx⟨f⟩=∫f(x)P(x)dx
evaluated over the limits of the probability density function, i.e. if your PDF runs from−∞−∞ to ∞∞, then those are your limits of integration.
In our case, the PDF is the Maxwell-Boltzmann distribution (denoted P(v)P(v)here) and the quantity we want to find the expectation value of is simply the velocity,vv. Since velocity can only be positive (as far as our model is concerned), the limits of integration are from 00 to ∞∞.
Therefore, the mean velocity ⟨v⟩⟨v⟩ is given by the integral
⟨v⟩=∫∞0vP(v)dv⟨v⟩=∫0∞vP(v)dv
where P(v)P(v) is the Maxwell-Boltzmann distribution
P(v)=(m2πkT)3/24πv2exp(−mv22kT)P(v)=(m2πkT)3/24πv2exp(−mv22kT)
So:
⟨v⟩=∫∞0vP(v)dv=4π(m2πkT)3/2∫∞0v3exp(−m2kTv2)dv⟨v⟩=∫0∞vP(v)dv=4π(m2πkT)3/2∫0∞v3exp(−m2kTv2)dv
The integral can be evaluated using integration by parts repeatedly. The process is not interesting and you could consult a table of standard integrals to find the result:
∫∞0v3exp(−αv2)dv=12α2∫0∞v3exp(−αv2)dv=12α2
Setting α=m/2kTα=m/2kT,
⟨v⟩=4π(m2πkT)3/24k2T22m2=8kTπm−−−−√⟨v⟩=4π(m2πkT)3/24k2T22m2=8kTπm
mm here refers to the mass of one molecule, whereas MM in your question refers to the molar mass of the compound. They are related by M=NAmM=NAm, where NANA is the Avogadro constant. Since R=NAkR=NAk, you can multiply top and bottom by NANA to obtain the desired result
⟨v⟩=8RTπM−−−−−√⟨v⟩=8RTπM
......
if you can't understand then follow this link
https://chemistry.stackexchange.com/questions/42253/derivation-of-mean-speed-from-maxwell-boltzmann-distribution
If one has a probability density function P(x)P(x), then the expectation value of a quantity f(x)f(x) is given by
⟨f⟩=∫f(x)P(x)dx⟨f⟩=∫f(x)P(x)dx
evaluated over the limits of the probability density function, i.e. if your PDF runs from−∞−∞ to ∞∞, then those are your limits of integration.
In our case, the PDF is the Maxwell-Boltzmann distribution (denoted P(v)P(v)here) and the quantity we want to find the expectation value of is simply the velocity,vv. Since velocity can only be positive (as far as our model is concerned), the limits of integration are from 00 to ∞∞.
Therefore, the mean velocity ⟨v⟩⟨v⟩ is given by the integral
⟨v⟩=∫∞0vP(v)dv⟨v⟩=∫0∞vP(v)dv
where P(v)P(v) is the Maxwell-Boltzmann distribution
P(v)=(m2πkT)3/24πv2exp(−mv22kT)P(v)=(m2πkT)3/24πv2exp(−mv22kT)
So:
⟨v⟩=∫∞0vP(v)dv=4π(m2πkT)3/2∫∞0v3exp(−m2kTv2)dv⟨v⟩=∫0∞vP(v)dv=4π(m2πkT)3/2∫0∞v3exp(−m2kTv2)dv
The integral can be evaluated using integration by parts repeatedly. The process is not interesting and you could consult a table of standard integrals to find the result:
∫∞0v3exp(−αv2)dv=12α2∫0∞v3exp(−αv2)dv=12α2
Setting α=m/2kTα=m/2kT,
⟨v⟩=4π(m2πkT)3/24k2T22m2=8kTπm−−−−√⟨v⟩=4π(m2πkT)3/24k2T22m2=8kTπm
mm here refers to the mass of one molecule, whereas MM in your question refers to the molar mass of the compound. They are related by M=NAmM=NAm, where NANA is the Avogadro constant. Since R=NAkR=NAk, you can multiply top and bottom by NANA to obtain the desired result
⟨v⟩=8RTπM−−−−−√⟨v⟩=8RTπM
......
if you can't understand then follow this link
https://chemistry.stackexchange.com/questions/42253/derivation-of-mean-speed-from-maxwell-boltzmann-distribution
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