Answer
$P_f=4.6$atm
Work Step by Step
We know that $P_1V_1^{\gamma}=P_2V_2^{\gamma}$
$\implies P_2={P_1}(\frac{V_1}{V_2})^{\gamma}$...........................eq(1)
For diatomic gases with adiabatic expansion:
$\gamma=\frac{C_P}{C_V}=\frac{\frac{7}{2}R}{\frac{5}{2}R}=\frac{7}{5}$
Substituting the values of $P_1, V_1,V_2$ and $\gamma$ into eq(1), we get:
$ P_2={32}(\frac{1\times10^{-3}}{4\times10^{-3}})^{\frac{7}{5}}=4.6 atm$
Thus $P_2$ or $P_f=4.6$atm
