# Chapter 3 - Differentiation - 3.2 The Derivative as a Function - Exercises - Page 115: 60

$$0.74$$

#### Work Step by Step

We are given: $$v_{\mathrm{avg}}=\sqrt{\frac{8 R T}{\pi M}}$$ Since $$R= 8.31,\ \ \ \ M= 0.032$$ Then $$v_{\mathrm{avg}}=\sqrt{\frac{8 R T}{\pi M}}= v_{\mathrm{avg}}=\sqrt{\frac{8 (8.31)T}{\pi (0.032)}} = \sqrt{\frac{2077.5}{\pi}}T^{1/2}$$ Hence \begin{align*} \frac{dv_{\mathrm{avg}}}{dT}&=\frac{1}{2} \sqrt{\frac{2077.5}{\pi}}T^{-1/2}\\ \frac{dv_{\mathrm{avg}}}{dT}\bigg|_{T=300}&=\frac{1}{2} \sqrt{\frac{2077.5}{\pi}}(300)^{-1/2}\\ &\approx 0.74 \end{align*}

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