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)`.

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 Problem
We need to calculate the rotational kinetic energy (RKE) of oxygen gas (O₂) given its mass, pressure, and density.

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = Pressure (in pascals)
- \( V \) = Volume (in cubic meters)
- \( n \) = Number of moles
- \( R \) = Universal gas constant (\( 8.314 \, \text{J/(mol K)} \))
- \( T \) = Temperature (in kelvin)

### Step 3: Relate Density to Pressure and Temperature
The density (\( \rho \)) of a gas can be expressed as:
\[ \rho = \frac{m}{V} \]
Where \( m \) is the mass of the gas. Rearranging gives:
\[ V = \frac{m}{\rho} \]

Substituting this into the ideal gas law gives:
\[ P \left(\frac{m}{\rho}\right) = nRT \]
This can be rearranged to find \( n \):
\[ n = \frac{P \cdot m}{R \cdot T \cdot \rho} \]

### Step 4: Calculate the Rotational Kinetic Energy
For a diatomic gas like oxygen, the total degrees of freedom is 5 (3 translational + 2 rotational). According to the equipartition theorem, the energy associated with each degree of freedom is \( \frac{1}{2}kT \) or \( \frac{1}{2}RT \) per mole.

The rotational kinetic energy (RKE) can be expressed as:
\[ RKE = nRT \]
Substituting \( n \) from the previous step:
\[ RKE = \frac{P \cdot m}{R \cdot T \cdot \rho} \cdot RT \]
This simplifies to:
\[ RKE = \frac{P \cdot m}{\rho} \]

### Step 5: Substitute Values
Given:
- Mass (\( m \)) = 2 kg
- Pressure (\( P \)) = 1 atm = \( 1.013 \times 10^5 \, \text{Pa} \)
- Density (\( \rho \)) = \( 2.03 \times 10^5 \, \text{kg/m}^3 \)

Now substituting these values:
\[ RKE = \frac{1.013 \times 10^5 \, \text{Pa} \cdot 2 \, \text{kg}}{2.03 \times 10^5 \, \text{kg/m}^3} \]

### Step 6: Calculate the Result
Calculating the above expression:
\[ RKE = \frac{2.026 \times 10^5 \, \text{Pa} \cdot \text{kg}}{2.03 \times 10^5 \, \text{kg/m}^3} \]
\[ RKE = \frac{2.026}{2.03} \, \text{J} \]
\[ RKE \approx 1 \, \text{J} \]

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

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