The speed v of the molecules of mass m of an ideal gas obeys Maxwell's velocity distribution law at an equilibrium temperature T. Let (vx, vy, vz) denote the components of the velo... The speed v of the molecules of mass m of an ideal gas obeys Maxwell's velocity distribution law at an equilibrium temperature T. Let (vx, vy, vz) denote the components of the velocity and kb the Boltzmann constant. The average value of (αv - βv), where α and β are constants, is: (a) (α2 - β2)kbT / m (b) (α2 + β2)kbT / m (c) (α + β)kbT / m. Choose the correct option.
Understand the Problem
The question is asking for the average value of a specific expression involving the components of velocity and constants related to an ideal gas obeying Maxwell's velocity distribution law at equilibrium temperature.
Answer
The average value is: $ \langle (\alpha v_x - \beta v_y) \rangle = (\alpha - \beta) \frac{k_B T}{m} $
Answer for screen readers
The average value is given by: $$ \langle (\alpha v_x - \beta v_y) \rangle = (\alpha - \beta) \frac{k_B T}{m} $$
Steps to Solve
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Understanding the Average Value
We need to find the average value of the expression $(\alpha v_x - \beta v_y)$. According to the problem, the averages of the velocity components for an ideal gas obeying Maxwell's distribution relate to temperature. -
Using the Average Velocity Values
From Maxwell's distribution, we know: $$ \langle v_x \rangle = \langle v_y \rangle = \sqrt{\frac{k_B T}{m}} $$ Thus, we can express the averages as: $$ \langle \alpha v_x \rangle = \alpha \langle v_x \rangle = \alpha \sqrt{\frac{k_B T}{m}} $$ and $$ \langle \beta v_y \rangle = \beta \langle v_y \rangle = \beta \sqrt{\frac{k_B T}{m}} $$ -
Combining the Averages
Now, we can find the average of the entire expression: $$ \langle \alpha v_x - \beta v_y \rangle = \alpha \langle v_x \rangle - \beta \langle v_y \rangle $$ Using the previous results: $$ \langle \alpha v_x - \beta v_y \rangle = \alpha \sqrt{\frac{k_B T}{m}} - \beta \sqrt{\frac{k_B T}{m}} $$ -
Factoring Out Common Terms
We can factor out the common term $\sqrt{\frac{k_B T}{m}}$: $$ \langle \alpha v_x - \beta v_y \rangle = \left(\alpha - \beta\right) \sqrt{\frac{k_B T}{m}} $$ -
Identifying the Correct Answer
After evaluating, we see that the result matches option (b): $$ \langle (\alpha v_x - \beta v_y) \rangle = (\alpha - \beta) \frac{k_B T}{m} $$
The average value is given by: $$ \langle (\alpha v_x - \beta v_y) \rangle = (\alpha - \beta) \frac{k_B T}{m} $$
More Information
The average value derived above relates to the behavior of molecules in an ideal gas, which behaves consistently under Maxwell's distribution at thermal equilibrium. This shows how macroscopic properties (like temperature) mediate the behavior of particles at the microscopic level.
Tips
- Incorrectly applying average values: Make sure to clearly define averages used in calculations.
- Forgetting temperature dependence: It's crucial to remember that average velocity is temperature-dependent in an ideal gas.
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