An ideal monoatomic gas is initially in state `1` with pressure `p_(1) = 20 atm` and volume `v_(1) = 1500 cm^(3)`. If is then taken to state `2` with pressure `p_(2) = 1.5 p_(1)` and volume `v_(2) = 2v_(1)`. The change in internal energy from state `1` to state `2` is equal to

Options

A. `2000J`
B. `3000 J`
C. `6000 J`
D. `9000 J`

Explanation

To solve the problem of finding the change in internal energy from state 1 to state 2 for an ideal monoatomic gas, we can follow these steps:

### Step 1: Understand the Internal Energy Formula
The change in internal energy (ΔU) for an ideal gas is given by the formula:
\[ \Delta U = N C_v \Delta T \]
where:
- \( N \) is the number of moles of the gas,
- \( C_v \) is the molar heat capacity at constant volume,
- \( \Delta T \) is the change in temperature.

For a monoatomic ideal gas, the value of \( C_v \) is:
\[ C_v = \frac{3}{2} R \]
where \( R \) is the universal gas constant.

### Step 2: Use the Ideal Gas Law
The ideal gas law states:
\[ PV = nRT \]
From this equation, we can derive the change in internal energy in terms of pressure and volume.

### Step 3: Express ΔU in terms of P and V
We can express the change in internal energy as:
\[ \Delta U = \frac{3}{2} N R \Delta T = \frac{3}{2} N R \left( \frac{\Delta (PV)}{R} \right) = \frac{3}{2} \Delta (PV) \]
Thus, we can write:
\[ \Delta U = \frac{3}{2} (P_f V_f - P_i V_i) \]
where:
- \( P_f \) and \( V_f \) are the final pressure and volume,
- \( P_i \) and \( V_i \) are the initial pressure and volume.

### Step 4: Substitute the Given Values
From the problem:
- \( P_1 = 20 \, \text{atm} \)
- \( V_1 = 1500 \, \text{cm}^3 \)
- \( P_2 = 1.5 P_1 = 1.5 \times 20 = 30 \, \text{atm} \)
- \( V_2 = 2 V_1 = 2 \times 1500 = 3000 \, \text{cm}^3 \)

Now, substituting these values into the equation for ΔU:
\[ \Delta U = \frac{3}{2} \left( P_2 V_2 - P_1 V_1 \right) \]
\[ = \frac{3}{2} \left( 30 \, \text{atm} \times 3000 \, \text{cm}^3 - 20 \, \text{atm} \times 1500 \, \text{cm}^3 \right) \]

### Step 5: Calculate the Values
Convert the units:
- \( 1 \, \text{atm} = 10^5 \, \text{N/m}^2 \)
- \( 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \)

Thus, we convert:
- \( P_1 = 20 \times 10^5 \, \text{N/m}^2 \)
- \( P_2 = 30 \times 10^5 \, \text{N/m}^2 \)
- \( V_1 = 1500 \times 10^{-6} \, \text{m}^3 \)
- \( V_2 = 3000 \times 10^{-6} \, \text{m}^3 \)

Now substituting these into the ΔU equation:
\[ \Delta U = \frac{3}{2} \left( (30 \times 10^5)(3000 \times 10^{-6}) - (20 \times 10^5)(1500 \times 10^{-6}) \right) \]
\[ = \frac{3}{2} \left( 30 \times 3 - 20 \times 1.5 \right) \times 10^5 \times 10^{-6} \]
\[ = \frac{3}{2} \left( 90 - 30 \right) \times 10^{-1} \]
\[ = \frac{3}{2} \times 60 \times 10^{-1} \]
\[ = 90 \times 10^{-1} = 9000 \, \text{J} \]

### Final Answer
The change in internal energy from state 1 to state 2 is:
\[ \Delta U = 9000 \, \text{J} \]