One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_(0)^(3))]` where `P_(0)` and `V_(0)` are constant. The translational kinetic energy of the gas when `V = V_(0)` is given by

Options

A. `5 P_(0) V_(0)//4`
B. `3 P_(0) V_(0)//4`
C. `3 P_(0) V_(0)//2`
D. `5 P_(0) V_(0)//2`

Explanation

To find the translational kinetic energy of one mole of a diatomic gas when \( V = V_0 \), we can follow these steps:

### Step 1: Substitute \( V = V_0 \) into the pressure equation
We are given the equation for pressure:
\[ P = \frac{P_0}{1 + \left(\frac{V}{V_0}\right)^3} \]
Now, substitute \( V = V_0 \):
\[ P = \frac{P_0}{1 + \left(\frac{V_0}{V_0}\right)^3} = \frac{P_0}{1 + 1} = \frac{P_0}{2} \]

### Step 2: Use the formula for translational kinetic energy
The translational kinetic energy (TKE) for an ideal gas is given by the formula:
\[ TKE = \frac{3}{2} PV \]
Now substitute the values we have found for \( P \) and \( V \):
\[ TKE = \frac{3}{2} \left(\frac{P_0}{2}\right) V_0 \]

### Step 3: Simplify the expression
Now we can simplify the expression:
\[ TKE = \frac{3}{2} \cdot \frac{P_0}{2} \cdot V_0 = \frac{3}{4} P_0 V_0 \]

### Conclusion
Thus, the translational kinetic energy of the gas when \( V = V_0 \) is:
\[ \boxed{\frac{3}{4} P_0 V_0} \]

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