Answer
$$167.43 ~m/(s*atm)$$
Work Step by Step
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:
$$\frac{dv}{dP} \bigg|_{P=1.5}=\frac{dv}{dP} \bigg|_{T=200}= \frac{2900}{\sqrt{300}}\approx 167.43 ~m/(s*atm)$$
