Let us start with the more abstract mathematical formalism of the statistical mechanics by introducing some new concepts. Let us start with defining $P_i$ is the probability of finding the system in some specific state "i". It is not the combination of outcomes for a particular event but the probability of a single outcome i.e. "i" refers to one particular microstate with energy $E_i$
So, the probability for the system to be found in this state is, $$ P_i = C\, N(E_0 - E_i) = C\, N(E)$$ where $ E_i + E = const. = E_0$ You may ask why we don't treat this event as the combination of two independent events and write the probability, as we did previous in two systems S,S' with energy E, E' as P(E) = C N(E) N(E')
because now we consider the first one as a single microstate.
Now, we assume E act as a reservoir for $E_i$, and it gives, on Taylor expansion and with suitable approximation, $$ \ln{N(E_0 - E_i)} = \ln{N(E_0)} - \beta{E_i} \\ N(E_0-E_i)= N(E_0) e^{-\beta{E_i}}$$ or simply, $$ P_i = C e^{-\beta{E_i}} = \sum_E N(E) e^{-\beta{E}}$$ We know in addition, the sum of all probabilities should give unity, $$ \sum_i P_i = C \sum_i e^{-\beta{E_i}} = 1 $$ or $$ C = \frac{1}{\sum_i e^{-\beta{E_i}}} \\ \rightarrow\,\,\,\,\,\, P_i = \frac{e^{-\beta{E_i}}}{\sum_i e^{-\beta{E_i}}}$$ Note that, the probabiliy depends exponentially on the value of energy E.
Defining a new term called Partition function and denoting with a new symbol, $$ X = \sum_i e^{-\beta{E_i}} $$ we try to express all other variable using this X,
The average of Energy is given by, $$ \langle{E}\rangle = \frac{\sum_i e^{-\beta{E_i}E_i}}{\sum_i e^{-\beta{E_i}}} $$ but using X, $$ \sum_i e^{-\beta{E_i}}E_i = -\frac{\partial{X}}{\partial{\beta}}$$ and $$ \langle{E}\rangle = -\frac{1}{X} \frac{\partial{\ln{X}}}{\partial\beta}$$ The generalized force is given by (similar procedure), $$ f = \frac{1}{\beta} \frac{\partial\ln{X}}{\partial{x}}$$ with the same notations used in the previous post. In particular, f = p when x = V. When we talk about pressure, we mean the average pressure.
Since, X is function of both $\beta$ and x, we get $$d(\ln{X}) = \beta{dW} - E d\beta \\ d\left(k\ln{X} + k\beta{E}\right) = \frac{dQ}{T} \\\rightarrow\,\,\,\,\,\, S = k\left(\ln{X} + \beta{E}\right)$$ where S - entropy.
The partition for an ideal gas can be calculated as, $$ X = \frac{V^N}{h^{3N}} \left[\int_{-\infty}^{\infty}e^{-\beta{\frac{p^2}{2m}}} \,\,d^3p_i\right]^{3N} $$ The gaussian integral gives, $$\int_{-\infty}^{\infty} e^{-\alpha{x^2}} \,dx = \sqrt{\frac{\pi}{\alpha}} $$
So, $$\int_{-\infty}^{\infty} e^{-\frac{\beta}{2m}p^2}\,dp = \sqrt{\frac{2m\pi}{\beta}}$$
And so, $$ X = V^N \sqrt{\frac{2\pi{m}}{\beta{h^2}}}^{3N} \\ \ln{X} = N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}\right] $$
And the mean energy is calculated to be $$ E = \frac{3}{2}kT$$
with this, the Entropy is calculated to be, $$ S = k (\ln{X}+\beta{E}) = k N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}+ \frac{3}{2}\right] \\~\\= kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T} + \frac{3}{2}\right]$$ But, it has some small correction due to Gibbs paradox, and it is corrected as, $$ X' = \frac{X}{N!} $$ So that, our equation will get modified to (using sterling's approximation), $$ \ln{X'} = \ln{X} - \ln{N!} = \ln{X} - N \ln{N} + N $$ $$ S = kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T}+\frac{3}{2} - \ln{N} + 1 \right] \\~\\= kN \left[\ln{\frac{V}{N}} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) + \frac{3}{2} \ln{T} + \frac{5}{2}\right]$$
which is the final result for Entropy.
Now, we can calculate Maxwell's velocity distribution law using these ideas and by assuming the molecules to be non-interacting classical ideal particles.
The probability of finding the particle in the position range r and r+dr and momentum range p and p+dp is given by the canonical distribution as we used in the partition function, we can convert it into position and velocity and obtain equation, $$ M(r,v) d^3r \,d^3v = C e^{-\frac{\beta{mv^2}}{2}} d^3r \, d^3v $$ and integrating through all over the space and for all velocity range, we get the total number of molecules , $$ N = C V \left[\int_{-\infty}^{\infty} e^{-\frac{\beta{mv_i^2}}{2}}dv_i \right]^3 \\~\\= CV \left(\frac{2\pi}{\beta{m}}\right)^{\frac{3}{2}} = N $$
And so,we get the value of C , and the Maxwell's velocity function (as it does depend only on "v"), $$ M(v) \,d^3r\,d^3v = \frac{N}{V} \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} e^{-\frac{mv^2}{2kT}} \,d^3r\,d^3v $$ Once we get this, we can ask for the various ways of velocity distribution.
We can calculate the old school stuff - the three kinds of velocities - but now with the correct mathematical formalism.
Average velocity: To calculate the average of anything, first we need the probability density function. With our equation, the density function is calculated by taking small volume in the velocity space, $$ m(v) dv = 4\pi{M(v)} v^2 dv $$ and the function should be normalized, and so, $$ \int_0^\infty m(v) dv = \frac{N}{V}$$
gives the mean velocity as, $$ \langle{v}\rangle = \frac{V}{N} \int_0^\infty m(v) v dv = \frac{4V\pi}{N} \int_0^\infty M(v) v^3 dv$$
Applying the value of M(v), we get, $$ \langle {v}\rangle = 4\pi \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} \int_0^\infty e^{-\frac{mv^2}{2kT}}v^3 \,dv = \sqrt{\frac{8kT}{\pi{m}}}$$
Similarly, the mean square speed is $$ \langle{v^2}\rangle = \sqrt{\frac{3kT}{m}}$$
Finally the most probable speed is calculated by finding the maximum of the function m(v). $$ \frac{dm}{dv} = 0 $$
Apart from the constants, $$ m(v) = c v^2 e^{-\frac{mv^2}{2kT}} = kv^2 e^{(\,\,)} $$ and so the maximum condition gives, $$ 2v e^{(\,\, )} - \frac{m}{kT} v^3 e^{(\,\,)} = 0$$ $$ v_{(mp)} = most \,\,probable\,\,speed\,\,=\sqrt{\frac{2kT}{m}} $$
Specifically, if the particles are moving in one dimension (let us say "x"), then we get, $$\frac{dm_x}{dx} = 0 $$
In one dimension, the factor $4\pi{v^2}$ would not come in the equation of m(v), so it is just $$ m(v) = c e^{-mv^2}{2kT} $$ and maximum condition gives, $$ c \left(-\frac{m}{kT}\right) v = 0 \\\rightarrow \,\,\,\,\,\, v_{(mp)} = 0 $$
So, the most probable speed and also the mean speed in one dimension is calculated to be zero. Since, the distribution curves attain maximum at v = 0 and it is also symmetrical about v = 0.
So, the probability for the system to be found in this state is, $$ P_i = C\, N(E_0 - E_i) = C\, N(E)$$ where $ E_i + E = const. = E_0$ You may ask why we don't treat this event as the combination of two independent events and write the probability, as we did previous in two systems S,S' with energy E, E' as P(E) = C N(E) N(E')
because now we consider the first one as a single microstate.
Now, we assume E act as a reservoir for $E_i$, and it gives, on Taylor expansion and with suitable approximation, $$ \ln{N(E_0 - E_i)} = \ln{N(E_0)} - \beta{E_i} \\ N(E_0-E_i)= N(E_0) e^{-\beta{E_i}}$$ or simply, $$ P_i = C e^{-\beta{E_i}} = \sum_E N(E) e^{-\beta{E}}$$ We know in addition, the sum of all probabilities should give unity, $$ \sum_i P_i = C \sum_i e^{-\beta{E_i}} = 1 $$ or $$ C = \frac{1}{\sum_i e^{-\beta{E_i}}} \\ \rightarrow\,\,\,\,\,\, P_i = \frac{e^{-\beta{E_i}}}{\sum_i e^{-\beta{E_i}}}$$ Note that, the probabiliy depends exponentially on the value of energy E.
Defining a new term called Partition function and denoting with a new symbol, $$ X = \sum_i e^{-\beta{E_i}} $$ we try to express all other variable using this X,
The average of Energy is given by, $$ \langle{E}\rangle = \frac{\sum_i e^{-\beta{E_i}E_i}}{\sum_i e^{-\beta{E_i}}} $$ but using X, $$ \sum_i e^{-\beta{E_i}}E_i = -\frac{\partial{X}}{\partial{\beta}}$$ and $$ \langle{E}\rangle = -\frac{1}{X} \frac{\partial{\ln{X}}}{\partial\beta}$$ The generalized force is given by (similar procedure), $$ f = \frac{1}{\beta} \frac{\partial\ln{X}}{\partial{x}}$$ with the same notations used in the previous post. In particular, f = p when x = V. When we talk about pressure, we mean the average pressure.
Since, X is function of both $\beta$ and x, we get $$d(\ln{X}) = \beta{dW} - E d\beta \\ d\left(k\ln{X} + k\beta{E}\right) = \frac{dQ}{T} \\\rightarrow\,\,\,\,\,\, S = k\left(\ln{X} + \beta{E}\right)$$ where S - entropy.
The partition for an ideal gas can be calculated as, $$ X = \frac{V^N}{h^{3N}} \left[\int_{-\infty}^{\infty}e^{-\beta{\frac{p^2}{2m}}} \,\,d^3p_i\right]^{3N} $$ The gaussian integral gives, $$\int_{-\infty}^{\infty} e^{-\alpha{x^2}} \,dx = \sqrt{\frac{\pi}{\alpha}} $$
So, $$\int_{-\infty}^{\infty} e^{-\frac{\beta}{2m}p^2}\,dp = \sqrt{\frac{2m\pi}{\beta}}$$
And so, $$ X = V^N \sqrt{\frac{2\pi{m}}{\beta{h^2}}}^{3N} \\ \ln{X} = N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}\right] $$
And the mean energy is calculated to be $$ E = \frac{3}{2}kT$$
with this, the Entropy is calculated to be, $$ S = k (\ln{X}+\beta{E}) = k N \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{m}}{h^2}) - \frac{3}{2} \ln{\beta}+ \frac{3}{2}\right] \\~\\= kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T} + \frac{3}{2}\right]$$ But, it has some small correction due to Gibbs paradox, and it is corrected as, $$ X' = \frac{X}{N!} $$ So that, our equation will get modified to (using sterling's approximation), $$ \ln{X'} = \ln{X} - \ln{N!} = \ln{X} - N \ln{N} + N $$ $$ S = kN \left[\ln{V} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) - \frac{3}{2} \ln{T}+\frac{3}{2} - \ln{N} + 1 \right] \\~\\= kN \left[\ln{\frac{V}{N}} + \frac{3}{2} \ln(\frac{2\pi{mk}}{h^2}) + \frac{3}{2} \ln{T} + \frac{5}{2}\right]$$
which is the final result for Entropy.
Now, we can calculate Maxwell's velocity distribution law using these ideas and by assuming the molecules to be non-interacting classical ideal particles.
The probability of finding the particle in the position range r and r+dr and momentum range p and p+dp is given by the canonical distribution as we used in the partition function, we can convert it into position and velocity and obtain equation, $$ M(r,v) d^3r \,d^3v = C e^{-\frac{\beta{mv^2}}{2}} d^3r \, d^3v $$ and integrating through all over the space and for all velocity range, we get the total number of molecules , $$ N = C V \left[\int_{-\infty}^{\infty} e^{-\frac{\beta{mv_i^2}}{2}}dv_i \right]^3 \\~\\= CV \left(\frac{2\pi}{\beta{m}}\right)^{\frac{3}{2}} = N $$
And so,we get the value of C , and the Maxwell's velocity function (as it does depend only on "v"), $$ M(v) \,d^3r\,d^3v = \frac{N}{V} \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} e^{-\frac{mv^2}{2kT}} \,d^3r\,d^3v $$ Once we get this, we can ask for the various ways of velocity distribution.
We can calculate the old school stuff - the three kinds of velocities - but now with the correct mathematical formalism.
Average velocity: To calculate the average of anything, first we need the probability density function. With our equation, the density function is calculated by taking small volume in the velocity space, $$ m(v) dv = 4\pi{M(v)} v^2 dv $$ and the function should be normalized, and so, $$ \int_0^\infty m(v) dv = \frac{N}{V}$$
gives the mean velocity as, $$ \langle{v}\rangle = \frac{V}{N} \int_0^\infty m(v) v dv = \frac{4V\pi}{N} \int_0^\infty M(v) v^3 dv$$
Applying the value of M(v), we get, $$ \langle {v}\rangle = 4\pi \left(\frac{m}{2\pi{kT}}\right)^{\frac{3}{2}} \int_0^\infty e^{-\frac{mv^2}{2kT}}v^3 \,dv = \sqrt{\frac{8kT}{\pi{m}}}$$
Similarly, the mean square speed is $$ \langle{v^2}\rangle = \sqrt{\frac{3kT}{m}}$$
Finally the most probable speed is calculated by finding the maximum of the function m(v). $$ \frac{dm}{dv} = 0 $$
Apart from the constants, $$ m(v) = c v^2 e^{-\frac{mv^2}{2kT}} = kv^2 e^{(\,\,)} $$ and so the maximum condition gives, $$ 2v e^{(\,\, )} - \frac{m}{kT} v^3 e^{(\,\,)} = 0$$ $$ v_{(mp)} = most \,\,probable\,\,speed\,\,=\sqrt{\frac{2kT}{m}} $$
Specifically, if the particles are moving in one dimension (let us say "x"), then we get, $$\frac{dm_x}{dx} = 0 $$
In one dimension, the factor $4\pi{v^2}$ would not come in the equation of m(v), so it is just $$ m(v) = c e^{-mv^2}{2kT} $$ and maximum condition gives, $$ c \left(-\frac{m}{kT}\right) v = 0 \\\rightarrow \,\,\,\,\,\, v_{(mp)} = 0 $$
So, the most probable speed and also the mean speed in one dimension is calculated to be zero. Since, the distribution curves attain maximum at v = 0 and it is also symmetrical about v = 0.
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