Find the rotational kinetic energy of 2 kg of oxygen gas molecules (in joules) at 1 atm pressure. The density of `O_(2)` is `2.03xx10^(5)kg//m^(3)`.

Options

A. 1J
B. 2J
C. 3J
D. 4J

Answer

To find the rotational kinetic energy of 2 kg of oxygen gas molecules at 1 atm pressure, we can follow these steps:

### Step 1: Understand the formula for rotational kinetic energy
The formula for rotational kinetic energy (KE_rot) in terms of the degrees of freedom (f), number of moles (n), the gas constant (R), and temperature (T) is given by:
\[ KE_{rot} = \frac{f}{2} nRT \]

### Step 2: Identify the degrees of freedom for oxygen
For diatomic gases like oxygen (O₂), the degrees of freedom related to rotation (f_rot) is 2. Therefore:
\[ f_{rot} = 2 \]

### Step 3: Use the ideal gas law to express nRT
From the ideal gas law, we know:
\[ PV = nRT \]
Where:
- P = pressure (in Pascals)
- V = volume
- n = number of moles
- R = universal gas constant (approximately \(8.314 \, J/(mol \cdot K)\))
- T = temperature (in Kelvin)

We can express \(nRT\) as:
\[ nRT = PV \]

### Step 4: Calculate the volume of the gas
The volume (V) can be expressed in terms of mass (m) and density (ρ):
\[ V = \frac{m}{\rho} \]

### Step 5: Substitute values into the equation
Now substituting \(V\) into the equation for \(nRT\):
\[ nRT = P \left(\frac{m}{\rho}\right) \]

### Step 6: Substitute into the rotational kinetic energy formula
Now substituting this into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{f_{rot}}{2} \cdot \frac{P \cdot m}{\rho} \]

### Step 7: Insert the known values
Given:
- Mass (m) = 2 kg
- Pressure (P) = 1 atm = \(1.013 \times 10^5 \, Pa\)
- Density (ρ) = \(2.03 \times 10^5 \, kg/m^3\)

Now substituting these values:
\[ KE_{rot} = \frac{2}{2} \cdot \frac{(1.013 \times 10^5) \cdot 2}{2.03 \times 10^5} \]

### Step 8: Calculate the rotational kinetic energy
Calculating the above expression:
\[ KE_{rot} = 1 \cdot \frac{(1.013 \times 10^5) \cdot 2}{2.03 \times 10^5} \]
\[ KE_{rot} = \frac{2.026 \times 10^5}{2.03 \times 10^5} \]
\[ KE_{rot} \approx 1 \, J \]

### Final Answer
The rotational kinetic energy of 2 kg of oxygen gas molecules at 1 atm pressure is approximately **1 Joule**.

---