A cubical container having each sides as `l` is filled with a gas having `N` molecules in the container. Mass of each molecule is `m`. If we assume that at every instant half of the molecules are moving towards the positive x-axis and half of the molecules are moving towards the negative x-axis.Two walls of the container are perpendicular to the x-axis. Find the net force acting on the two walls given? Assume that all the molecules are moving with speed `v_(0)`.

Answer

To solve the problem, we need to find the net force acting on the two walls of a cubical container filled with gas molecules. Let's break down the solution step by step.

### Step 1: Understanding the Setup
We have a cubical container with side length \( l \) filled with \( N \) gas molecules, each having a mass \( m \). Half of the molecules are moving towards the positive x-axis and half towards the negative x-axis. The speed of each molecule is \( v_0 \).

### Step 2: Calculate the Momentum Change
When a molecule collides with a wall, it exerts a force on that wall. The change in momentum of a single molecule when it collides with the wall can be calculated as:
- Momentum before collision (moving towards the wall): \( p_{\text{before}} = mv_0 \)
- Momentum after collision (moving away from the wall): \( p_{\text{after}} = -mv_0 \)
- Change in momentum \( \Delta p = p_{\text{after}} - p_{\text{before}} = -mv_0 - mv_0 = -2mv_0 \)

### Step 3: Calculate the Number of Collisions
Since half of the \( N \) molecules are moving towards the wall, the number of molecules hitting one wall per unit time can be calculated. The time between collisions for a molecule traveling at speed \( v_0 \) to travel a distance \( l \) (the distance to the wall) is:
\[ \Delta t = \frac{l}{v_0} \]
Thus, the number of molecules hitting the wall in this time is:
\[ N_A = \frac{N}{2} \cdot \frac{v_0}{l} \]

### Step 4: Calculate the Force on One Wall
The force exerted on one wall due to the molecules colliding can be calculated using the rate of change of momentum:
\[ F_A = N_A \cdot \Delta p = \left(\frac{N}{2} \cdot \frac{v_0}{l}\right) \cdot (-2mv_0) = -\frac{Nmv_0^2}{l} \]
Since force is a vector, we consider the magnitude:
\[ F_A = \frac{Nmv_0^2}{l} \]

### Step 5: Calculate the Force on the Opposite Wall
The force on the opposite wall (Wall B) will have the same magnitude but will act in the opposite direction:
\[ F_B = \frac{Nmv_0^2}{l} \]

### Step 6: Calculate the Net Force
The net force acting on the two walls is the sum of the forces acting on them:
\[ F_{\text{net}} = F_B - F_A = \frac{Nmv_0^2}{l} - \left(-\frac{Nmv_0^2}{l}\right) = \frac{Nmv_0^2}{l} + \frac{Nmv_0^2}{l} = \frac{2Nmv_0^2}{l} \]

### Final Answer
The net force acting on the two walls of the container is:
\[ F_{\text{net}} = \frac{2Nmv_0^2}{l} \]