`V = k((P)/(T))^(0.33)` where k is constant. It is an,

Options

A. isothermal process
B. adiabatic process
C. isochoric process
D. isobaric process

Explanation

To solve the problem, we need to analyze the given equation for volume \( V \) in terms of pressure \( P \) and temperature \( T \):

### Given:
\[ V = k \left( \frac{P}{T} \right)^{0.33} \]

where \( k \) is a constant.

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]

From this equation, we can express \( \frac{P}{T} \):
\[ \frac{P}{T} = \frac{nR}{V} \]

### Step 2: Substitute \( \frac{P}{T} \) in the Volume Equation
Now, we can substitute \( \frac{P}{T} \) into the volume equation:
\[ V = k \left( \frac{nR}{V} \right)^{0.33} \]

### Step 3: Simplify the Equation
Rearranging the equation gives:
\[ V = k (nR)^{0.33} \cdot V^{-0.33} \]

### Step 4: Multiply Both Sides by \( V^{0.33} \)
To eliminate the negative exponent, we multiply both sides by \( V^{0.33} \):
\[ V^{1 + 0.33} = k (nR)^{0.33} \]
\[ V^{1.33} = k (nR)^{0.33} \]

### Step 5: Introduce a New Constant
Since \( k \), \( n \), and \( R \) are all constants, we can introduce a new constant \( k' \):
\[ V^{1.33} = k' \]

### Step 6: Analyze the Result
This equation indicates that the volume \( V \) raised to the power of \( 1.33 \) is equal to a constant \( k' \). This implies that the volume does not change with respect to pressure and temperature, which is characteristic of an isochoric process (constant volume).

### Conclusion
Thus, the process described by the equation \( V = k \left( \frac{P}{T} \right)^{0.33} \) is an **isochoric process**.

### Final Answer:
The process is an **isochoric process**.

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