The given formula for the average speed $v_{\text{avg}}$ of a gas molecule is:

$v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}$

Where:

• $R = 8.31 \, \text{J/(mol·K)}$ (gas constant),
• $T$ is the temperature in kelvins,
• $M$ is the molar mass in kg/mol,
• $\pi$ is the mathematical constant (approximately $3.1416$).

We are tasked with finding $\frac{d}{dT}v_{\text{avg}}$, the derivative of $v_{\text{avg}}$ with respect to temperature $T$.

Step 1: Differentiate $v_{\text{avg}}$ with respect to $T$

The function $v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}$ can be rewritten as:

$v_{\text{avg}} = \left( \frac{8R}{\pi M} \right)^{1/2} T^{1/2}$

Now, take the derivative of $v_{\text{avg}}$ with respect to $T$:

$\frac{d}{dT} v_{\text{avg}} = \frac{1}{2} \left( \frac{8R}{\pi M} \right)^{1/2} T^{-1/2}$

Step 2: Substitute known values

Given:

• $T = 350 \, \text{K}$,
• $R = 8.31 \, \text{J/(mol·K)}$,
• $M = 0.032 \, \text{kg/mol}$,
• $\pi = 3.1416$.

Substituting these values into the derivative expression:

$\frac{d}{dT} v_{\text{avg}} = \frac{1}{2} \left( \frac{8 \times 8.31}{3.1416 \times 0.032} \right)^{1/2} \cdot 350^{-1/2}$

Let's calculate this step by step.The derivative of the average speed $v_{\text{avg}}$ with respect to temperature $T$ at $T = 350 \, \text{K}$ for oxygen is approximately:

$\frac{d}{dT} v_{\text{avg}} = 0.6873 \, \text{m/(s·K)}$

Would you like any further details or clarifications?

Here are 5 related questions to expand on this topic:

1. How does the molar mass affect the average speed of a gas molecule?
2. What is the physical significance of the derivative $\frac{d}{dT} v_{\text{avg}}$?
3. How does the average speed change if the temperature is doubled?
4. Can this formula be applied to real gases, or is it specific to ideal gases?
5. How does this average speed formula relate to the kinetic theory of gases?

Tip: When differentiating a square root function, express it as a fractional power for easier calculation.