A

Step 1: When the container suddenly stops, the gas inside will experience an inertial effect due to its initial velocity. The gas will continue moving with the velocity it had before the container stopped.
Step 2: Using the kinetic energy formula, we can express the kinetic energy of the gas before stopping as follows:
$$ KE = \frac{1}{2} mv^2 $$
where \( m \) is the mass of the gas and \( v \) is its velocity (30 m/s).
Step 3: The work done to bring the gas to rest is equal to the negative change in kinetic energy. This work done will lead to a change in internal energy, and considering the gas is monoatomic, the change in the internal energy is related to the change in temperature by:
$$ \Delta U = \frac{3}{2} n R \Delta T $$
Step 4: Setting the work done equal to the change in internal energy, we get:
$$ \frac{1}{2} mv^2 = \frac{3}{2} n R \Delta T $$
Step 5: Solving for \( \Delta T \):
$$ \Delta T = \frac{mv^2}{3nR} $$
Substituting the known values: \( m = 4.0u \), \( v = 30 \text{ m/s} \), and substituting \( R \) with its appropriate value gives:
Therefore, A.