Step 1: Understanding the Question:
The problem asks for the effect on the propagation speed of a sound wave in a gaseous medium when its external pressure is increased to twice its initial value, keeping the system temperature constant.
Step 2: Key Formula or Approach:
According to Laplace's correction, the velocity $v$ of a sound wave through an ideal gas is given by:
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
where $\gamma$ is the adiabatic index, $P$ is the pressure, and $\rho$ is the gas mass density. From the ideal gas equation, we can rewrite the pressure-to-density ratio using temperature:
$$PV = nRT = \frac{m}{M}RT \implies P = \frac{\rho RT}{M} \implies \frac{P}{\rho} = \frac{RT}{M}$$
Substituting this back into the velocity formula gives:
$$v = \sqrt{\frac{\gamma RT}{M}}$$
Step 3: Detailed Explanation:
The rewritten velocity formula $v = \sqrt{\frac{\gamma RT}{M}}$ demonstrates that the speed of sound in a gas is uniquely determined by its absolute temperature $T$ and molecular properties ($\gamma, M$).
When the external pressure is doubled at a constant temperature, Boyle's Law dictates that the volume of the gas shrinks to half, which directly doubles its density $\rho$.
Because both $P$ and $\rho$ increase by the exact same factor of 2, their ratio $\frac{P}{\rho}$ remains completely constant.
Consequently, altering the pressure has no physical effect on the overall wave propagation velocity as long as the temperature is kept constant.
Step 4: Final Answer:
The speed of sound remains completely the same, matching option (A).