start from the Maxwell-Boltzmann distribution and derive an expression for the most probable speed of a gas of molecules at a temperature T. Go on t demonstrate the validity of the equipartition conclusion that the average translational kinetic energy of molecules free to move in three dimensions is (3/2)KT.
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To derive the expression for the most probable speed of a gas of molecules at a temperature T using the Maxwell-Boltzmann distribution, we first need to understand the distribution itself.
The Maxwell-Boltzmann distribution describes the distribution of speeds of molecules in an ideal gas at a given temperature. The probability distribution function for the speed of a gas molecule is given by:
f(v) = (4πv²) * (m / (2πkT))^(3/2) * exp(-(mv² / 2kT))
Where:
- f(v) is the probability distribution function of speed v,
- m is the mass of the gas molecule,
- k is the Boltzmann constant (1.380649 x 10^-23 J/K),
- T is the temperature in Kelvin.
The most probable speed, denoted as vmp, corresponds to the peak of the Maxwell-Boltzmann distribution. To find it, we need to find the value of v that maximizes the probability distribution function f(v).
To find the maximum, we can take the derivative of f(v) with respect to v and set it equal to zero:
d(f(v)) / d(v) = 0
To simplify the calculation, it's common to work with the natural logarithm of f(v). So, we'll take the natural logarithm of f(v) and then differentiate it:
ln(f(v)) = ln(4πv²) + (3/2)ln(m / (2πkT)) - (mv² / 2kT)
Differentiating ln(f(v)):
d(ln(f(v))) / d(v) = d(ln(4πv²)) / d(v) + (3/2)d(ln(m / (2πkT))) / d(v) - (md(v²) / d(v) * (1 / 2kT)) - (v² / 2kT)d(m / (2πkT)) / d(v)
Simplifying the equation and setting it equal to zero:
0 = (d(ln(4πv²)) / d(v)) + (3/2)(d(ln(m / (2πkT))) / d(v)) - (mv² / 2kT) - (v² / 2kT)(d(m / (2πkT)) / d(v))
The first derivative term, d(ln(4πv²)) / d(v), becomes:
d(ln(4πv²)) / d(v) = 2v^-1
The second derivative term, d(ln(m / (2πkT))) / d(v), becomes:
d(ln(m / (2πkT))) / d(v) = 0
Since m and T are constants, their derivatives with respect to speed v are zero.
Simplifying the equation further:
0 = 2v^-1 + (3/2)(0) - (mv² / 2kT) - (v² / 2kT)(0)
0 = 2v^-1 - (mv² / 2kT)
Rearranging the equation:
2v^-1 = mv² / 2kT
Canceling out the common terms:
1/v = mv / kT
1 = v² / kT
v² = kT
v = √(kT / m)
Therefore, the expression for the most probable speed of a gas of molecules at temperature T is given by:
vmp = √(kT / m)
Now, let's move on to demonstrate the validity of the equipartition conclusion.
The equipartition theorem states that, in thermal equilibrium, each degree of freedom of a molecule contributes (1/2)kT to the average energy. For a molecule free to move in three dimensions, we have a total of three translational degrees of freedom.
The average translational kinetic energy of a molecule is given by:
<E> = (3 / 2) kT
Here, (3 / 2) represents the contribution from the three degrees of freedom, and kT is the average energy for each degree of freedom. Hence, the validity of the equipartition conclusion is confirmed.
Note: This is a simplified explanation and derivation. In practice, the Maxwell-Boltzmann distribution can involve more mathematical steps and considerations, but this explanation provides an overview of the process.