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Show that the rms speed of a molecule in an ideal gas at absolute temperature \(T\) is given by $$v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}$$ where \(k\) is Boltzmann's constant and \(m\) is the mass of a molecule.

Short Answer

Expert verified
Answer: The root-mean-square (rms) speed of a molecule in an ideal gas at absolute temperature T is given by the formula $$v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}$$, where k is Boltzmann's constant and m is the mass of the molecule.

Step by step solution

01

Understand the equipartition theorem

The equipartition theorem states that in thermal equilibrium, the energy of an ideal gas is equally distributed among all its independent degrees of freedom. In our case, the independent degrees of freedom are the three dimensions of motion (x, y, and z).
02

Relate kinetic energy to speed

We know that the kinetic energy of a molecule can be expressed as $$K=\frac{1}{2}mv^2$$ where m is the mass of the molecule and v is its speed.
03

Apply the equipartition theorem

According to the equipartition theorem, the average kinetic energy per molecule is given by $$\langle{K}\rangle=\frac{3}{2}kT$$ Here, \(\langle{K}\rangle\) is the average kinetic energy, k is the Boltzmann's constant, and T is the absolute temperature.
04

Find the average of the square of the speeds

Now, we know that the average kinetic energy is connected to the square of the speeds, as seen in step 2. So, we can write: $$\frac{1}{2}m\langle{v^2}\rangle=\frac{3}{2}kT$$ where \(\langle{v^2}\rangle\) represents the average value of the square of the speeds.
05

Solve for the rms speed

To find the rms speed, we take the square root of the average value of the square of the speeds, which gives $$v_{\mathrm{rms}}=\sqrt{\langle{v^2}\rangle}$$ Now, we can solve for \(v_{\mathrm{rms}}\) using the equation from step 4: $$v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}$$ We have shown that the rms speed of a molecule in an ideal gas at absolute temperature T is given by $$v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}$$

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