One mole of an ideal monoatomic gas is caused to go through the cycle shown in figure. Then the change in the internal energy of gas from a to b and b to c is respectively

Options

A. `(9P_@V_@)/2,6 RT_@`
B. `(9P_@V_@)/2, 10 RT_@`
C. `(15P_@V_@)/2,6 RT_@`
D. `(15P_@V_@)/2, 10 RT_@`

Explanation

To solve the problem of finding the change in internal energy of an ideal monoatomic gas as it goes through the cycle from A to B and from B to C, we can follow these steps:

### Step 1: Understand the Internal Energy Change Formula
The change in internal energy (ΔU) for an ideal gas can be calculated using the formula:
\[ \Delta U = \frac{f}{2} n R \Delta T \]
where:
- \( f \) is the degrees of freedom (for a monoatomic gas, \( f = 3 \)),
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( \Delta T \) is the change in temperature.

### Step 2: Analyze the Process from A to B
For the process from A to B:
- Since it is a monoatomic gas, \( f = 3 \).
- The change in internal energy can be expressed as:
\[ \Delta U_{A \to B} = \frac{3}{2} n R (T_B - T_A) \]
- We need to find the temperatures at points A and B using the ideal gas law \( PV = nRT \).

Assuming at point A:
- \( P_A = P_0 \)
- \( V_A = V_0 \)

Then, using the ideal gas law:
\[ P_0 V_0 = n R T_A \implies T_A = \frac{P_0 V_0}{nR} \]

At point B:
- \( P_B = P_0 \)
- \( V_B = 4V_0 \)

Using the ideal gas law again:
\[ P_0 (4V_0) = n R T_B \implies T_B = \frac{P_0 (4V_0)}{nR} \]

### Step 3: Calculate ΔU from A to B
Now, substituting the values of \( T_A \) and \( T_B \):
\[ \Delta U_{A \to B} = \frac{3}{2} n R \left( \frac{P_0 (4V_0)}{nR} - \frac{P_0 V_0}{nR} \right) \]
\[ = \frac{3}{2} n R \left( \frac{3P_0 V_0}{nR} \right) = \frac{9}{2} P_0 V_0 \]

### Step 4: Analyze the Process from B to C
For the process from B to C:
- At point C, let’s assume:
- \( P_C = 2P_0 \)
- \( V_C = 4V_0 \)

Using the ideal gas law:
\[ 2P_0 (4V_0) = n R T_C \implies T_C = \frac{8P_0 V_0}{nR} \]

Now, we can calculate ΔU from B to C:
\[ \Delta U_{B \to C} = \frac{3}{2} n R (T_C - T_B) \]
Substituting the values:
\[ \Delta U_{B \to C} = \frac{3}{2} n R \left( \frac{8P_0 V_0}{nR} - \frac{4P_0 V_0}{nR} \right) \]
\[ = \frac{3}{2} n R \left( \frac{4P_0 V_0}{nR} \right) = 6P_0 V_0 \]

### Step 5: Final Values
Thus, the changes in internal energy are:
- From A to B: \( \Delta U_{A \to B} = \frac{9}{2} P_0 V_0 \)
- From B to C: \( \Delta U_{B \to C} = 6P_0 V_0 \)

### Summary
The change in internal energy of the gas from A to B is \( \frac{9}{2} P_0 V_0 \) and from B to C is \( 6P_0 V_0 \).