Energy of all molecules of a monatomic gas having a volume `V` and pressure `P` is `3//2 PV`. The total translational kinetic energy of all molecules of a diatomic gas at the same volume and pressure is

Options

A. `1//2 PV`
B. `3//2 PV`
C. `5//2 PV`
D. `3 PV`

Explanation

To find the total translational kinetic energy of all molecules of a diatomic gas at the same volume \( V \) and pressure \( P \), we can follow these steps:

### Step 1: Understand the relationship for monatomic gas
For a monatomic gas, the total translational kinetic energy (TKE) is given by:
\[ \text{TKE}_{\text{mono}} = \frac{3}{2} PV \]
This formula is derived from the kinetic theory of gases, where the average kinetic energy per molecule is \(\frac{3}{2} kT\) and relates to the macroscopic quantities \(P\) (pressure) and \(V\) (volume).

### Step 2: Apply the ideal gas law
Using the ideal gas law:
\[ PV = nRT \]
where \(n\) is the number of moles and \(R\) is the universal gas constant. For one mole of gas, we can write:
\[ PV = RT \]

### Step 3: Determine the degrees of freedom for diatomic gas
For a diatomic gas, the degrees of freedom are greater than that of a monatomic gas. A diatomic molecule has 5 degrees of freedom (3 translational and 2 rotational). Thus, the total translational kinetic energy for a diatomic gas can be expressed as:
\[ \text{TKE}_{\text{diatomic}} = \frac{5}{2} nRT \]

### Step 4: Substitute \(PV\) in the expression for diatomic gas
Since we know from the ideal gas law that \(PV = nRT\), we can substitute \(nRT\) with \(PV\):
\[ \text{TKE}_{\text{diatomic}} = \frac{5}{2} PV \]

### Step 5: Conclusion
Thus, the total translational kinetic energy of all molecules of a diatomic gas at the same volume \(V\) and pressure \(P\) is:
\[ \text{TKE}_{\text{diatomic}} = \frac{5}{2} PV \]

### Final Answer
The total translational kinetic energy of all molecules of a diatomic gas at the same volume and pressure is \(\frac{5}{2} PV\).

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