Answer
$1.18\times10^{7}Pa$
Work Step by Step
Here we use equation 15.2 $W=P\Delta V$ to find the average pressure of the expanding gas,
$W=P\Delta V=>P=\frac{W}{\Delta V}=\frac{\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})}{\pi r^{2}L}$
Let's plug known values into this equation.
$P=\frac{\frac{1}{2}(2.6\times10^{-3}kg)[(370\space m/s)^{2}-0]}{\pi (2.8\times10^{-3})^{2}(0.61\space m)}=1.18\times10^{7}Pa$
The average pressure of the = $1.18\times10^{7}Pa$
expanding gas
