An ideal monoatomic gas initially in state 1 with pressure `P_(1)=20` atm and volume `V_(1)1500cm^(3)` it is then taken to state 2 with pressure `P_(2)=1.5P_(1)` and volume `V_(2)=2V_(1)` find the change in internal energy in this process in KJ. (take `1atm` lit `=100J`)

Explanation

To find the change in internal energy of the ideal monoatomic gas during the process from state 1 to state 2, we will follow these steps:

### Step 1: Identify the initial and final states
- Initial state (State 1):
- Pressure, \( P_1 = 20 \) atm
- Volume, \( V_1 = 1500 \) cm³

- Final state (State 2):
- Pressure, \( P_2 = 1.5 P_1 = 1.5 \times 20 = 30 \) atm
- Volume, \( V_2 = 2 V_1 = 2 \times 1500 = 3000 \) cm³

### Step 2: Use the formula for change in internal energy
The change in internal energy (\( \Delta E \)) for an ideal monoatomic gas is given by:
\[ \Delta E = N C_v \Delta T \]
Where:
- \( C_v \) for a monoatomic gas = \( \frac{3}{2} R \)
- \( \Delta T = T_2 - T_1 \)

### Step 3: Calculate temperatures \( T_1 \) and \( T_2 \)
Using the ideal gas law \( PV = nRT \), we can express temperatures as:
\[ T_1 = \frac{P_1 V_1}{nR} \]
\[ T_2 = \frac{P_2 V_2}{nR} \]

### Step 4: Substitute \( T_1 \) and \( T_2 \) into the equation for \( \Delta E \)
Substituting the expressions for \( T_1 \) and \( T_2 \) into the equation for \( \Delta E \):
\[ \Delta E = N C_v \left( \frac{P_2 V_2}{nR} - \frac{P_1 V_1}{nR} \right) \]
This simplifies to:
\[ \Delta E = N \frac{3}{2} R \left( \frac{P_2 V_2 - P_1 V_1}{nR} \right) \]
\[ \Delta E = N \frac{3}{2} \left( P_2 V_2 - P_1 V_1 \right) \]

### Step 5: Calculate \( P_2 V_2 \) and \( P_1 V_1 \)
Now, we calculate:
\[ P_1 V_1 = 20 \, \text{atm} \times 1500 \, \text{cm}^3 = 30000 \, \text{atm cm}^3 \]
\[ P_2 V_2 = 30 \, \text{atm} \times 3000 \, \text{cm}^3 = 90000 \, \text{atm cm}^3 \]

### Step 6: Find \( \Delta E \)
Now substituting these values:
\[ \Delta E = N \frac{3}{2} (90000 - 30000) = N \frac{3}{2} \times 60000 = N \times 90000 \]

### Step 7: Convert units
Since \( 1 \, \text{atm} \, \text{cm}^3 = 100 \, \text{J} \):
\[ \Delta E = 90000 \, \text{atm cm}^3 \times 100 \, \text{J/atm cm}^3 = 9000000 \, \text{J} \]
Convert to kilojoules:
\[ \Delta E = \frac{9000000 \, \text{J}}{1000} = 9000 \, \text{kJ} \]

### Final Answer
The change in internal energy in this process is \( 9 \, \text{kJ} \).

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