The total kinetic energy of translational motion of all the molecules of 5 L of nitrogen exerting a pressure P is 3000 J.

Options

A. The total kinetic energy of 10 L of `N_2` at a pressure of 2P is 3000 J.
B. The total kinetic energy of 10 L of He at a pressure of 2P is 3000 J.
C. The total kinetic energy of 10 L of `O_2` at a pressure of 2P is 20,000 J.
D. The total kinetic energy of 10 L of Ne at a pressure of 2P is 12,000 J.

Answer

To solve the problem, we need to find the pressure exerted by the nitrogen gas given its total kinetic energy and volume. We will use the formula for the total kinetic energy of a gas:

\[ KE = \frac{3}{2} P V \]

where:
- \( KE \) is the total kinetic energy,
- \( P \) is the pressure,
- \( V \) is the volume.

### Step 1: Identify the given values
- Total kinetic energy, \( KE = 3000 \, \text{J} \)
- Volume, \( V = 5 \, \text{L} = 5 \times 10^{-3} \, \text{m}^3 \) (conversion from liters to cubic meters)

### Step 2: Rearrange the formula to solve for pressure \( P \)
From the kinetic energy formula, we can rearrange it to find \( P \):

\[ P = \frac{2 \times KE}{3 \times V} \]

### Step 3: Substitute the known values into the equation
Now we substitute \( KE \) and \( V \) into the equation:

\[ P = \frac{2 \times 3000 \, \text{J}}{3 \times (5 \times 10^{-3} \, \text{m}^3)} \]

### Step 4: Calculate the pressure
Calculating the denominator:

\[ 3 \times (5 \times 10^{-3}) = 15 \times 10^{-3} = 0.015 \, \text{m}^3 \]

Now substituting back into the pressure equation:

\[ P = \frac{6000}{0.015} = 400000 \, \text{Pa} = 4 \times 10^5 \, \text{Pa} \]

### Final Result
The pressure \( P \) exerted by the nitrogen gas is:

\[ P = 4 \times 10^5 \, \text{Pa} \]