## Calculus (3rd Edition)

$$167.43 ~m/(s*atm)$$
Given $$v(T ) = 29\sqrt{T },\ \ \ \ \ T= 200P$$ Since $$\frac{dv}{dT}= \frac{29}{2\sqrt{T}} ,\ \ \ \frac{dT}{dP}=200$$ Then \begin{align*} \frac{dv}{dP}&= \frac{dv}{dT}\frac{dT}{dP}\\ &= \frac{2900}{\sqrt{T}} \end{align*} Since at $P=1.5,\ \ T= 300$, then we get: $$\frac{dv}{dP} \bigg|_{P=1.5}=\frac{dv}{dP} \bigg|_{T=200}= \frac{2900}{\sqrt{300}}\approx 167.43 ~m/(s*atm)$$