An ideal gaseous sample at initial state I `(P_0, V_0, T_0)` is allowed to expand to volume `2V_0` using two different process, in the first process the equation of process is `PV_2=K_1` and in second process the equation of the process is `PV=K_2`.Then

Options

A. Work done in first process will be greater than work in a second process (magnitude wise)
B. The order of values of work done can not be compared unless we know the value of `K_1` and `K_2`
C. Values of work done (magnitude) in second process in greater in above expansion irrespective of the value of `K_1` and `K_2`
D. `I^(st)` process is not possible

Explanation

To solve the problem, we need to analyze the two processes through which the ideal gas expands from an initial state \((P_0, V_0, T_0)\) to a final volume of \(2V_0\). The two processes are defined by the equations \(PV^2 = K_1\) and \(PV = K_2\). Let's break down the steps to find the work done in each process.

### Step 1: Understand the Processes
1. **Process 1**: \(PV^2 = K_1\)
- This indicates a non-linear relationship between pressure and volume.

2. **Process 2**: \(PV = K_2\)
- This indicates an isothermal process (constant temperature) since the product of pressure and volume is constant.

### Step 2: Determine the Work Done in Each Process
The work done \(W\) in a process can be calculated using the integral:
\[ W = \int_{V_i}^{V_f} P \, dV \]

#### For Process 1: \(PV^2 = K_1\)
From the equation, we can express pressure \(P\) as:
\[ P = \frac{K_1}{V^2} \]
Now, substituting this into the work integral:
\[ W_1 = \int_{V_0}^{2V_0} \frac{K_1}{V^2} \, dV \]
Calculating the integral:
\[ W_1 = K_1 \left[-\frac{1}{V}\right]_{V_0}^{2V_0} = K_1 \left(-\frac{1}{2V_0} + \frac{1}{V_0}\right) = K_1 \left(\frac{1}{2V_0}\right) = \frac{K_1}{2V_0} \]

#### For Process 2: \(PV = K_2\)
From the equation, we can express pressure \(P\) as:
\[ P = \frac{K_2}{V} \]
Now, substituting this into the work integral:
\[ W_2 = \int_{V_0}^{2V_0} \frac{K_2}{V} \, dV \]
Calculating the integral:
\[ W_2 = K_2 \left[\ln V\right]_{V_0}^{2V_0} = K_2 \left(\ln(2V_0) - \ln(V_0)\right) = K_2 \ln(2) \]

### Step 3: Compare the Work Done
Now we have:
- \(W_1 = \frac{K_1}{2V_0}\)
- \(W_2 = K_2 \ln(2)\)

To compare \(W_1\) and \(W_2\), we need to analyze the constants \(K_1\) and \(K_2\). Without specific values for \(K_1\) and \(K_2\), we cannot definitively compare the magnitudes of work done in the two processes.

### Conclusion
Based on the analysis:
- The work done in the first process cannot be compared to the work done in the second process without knowing the values of \(K_1\) and \(K_2\).

Thus, the correct answer is:
**Option B**: The order of value of work done cannot be compared unless we know the value of \(K_1\) and \(K_2\).