Step 1: Recall Poisson’s relation for adiabatic process.
\[
p V^\gamma = \text{constant}, \gamma = \frac{C_p}{C_v} = 1.41
\]
Step 2: Relating density and pressure.
Density $\rho = \frac{p}{RT}$.
For adiabatic process:
\[
T \propto p^{(\gamma-1)/\gamma}
\]
So,
\[
\rho \propto \frac{p}{T} \propto p^{1/\gamma}
\]
Step 3: Density condition.
If $\rho_2 = \tfrac{1}{2}\rho_1$:
\[
\frac{\rho_2}{\rho_1} = \left(\frac{p_2}{p_1}\right)^{1/\gamma} = \frac{1}{2}
\]
\[
\left(\frac{p_2}{1000}\right)^{1/1.41} = 0.5
\]
\[
\frac{p_2}{1000} = 0.5^{1.41} = 0.375
\]
\[
p_2 = 375 \, hPa
\]
Final Answer:
\[
\boxed{375.00 \, hPa}
\]