Two moles of an ideal monoatomic gas, initially at pressure `p_1` and volume `V_1`, undergo an adiabatic compression until its volume is `V_2`. Then the gas is given heat Q at constant volume `V_2`.
(i) Sketch the complete process on a p-V diagram.
(ii) Find the total work done by the gas, the total change in its internal energy and the final temperature of the gas. [Give your answer in terms of `p_1,V_1,V_2, Q and R`]

Explanation

To solve the problem step-by-step, we will break it down into parts as follows:

### Part (i): Sketch the complete process on a p-V diagram

1. **Identify the Initial State**: The initial state of the gas is given by pressure \( p_1 \) and volume \( V_1 \). Plot this point on the p-V diagram.

2. **Adiabatic Compression**: The gas undergoes adiabatic compression from volume \( V_1 \) to \( V_2 \). This process is represented by a curve on the p-V diagram, moving from point 1 (at \( V_1, p_1 \)) to point 2 (at \( V_2, p_2 \)). The pressure increases as the volume decreases.

3. **Constant Volume Heating**: After reaching volume \( V_2 \), the gas is heated at constant volume \( V_2 \). This process is represented by a vertical line moving from point 2 (at \( V_2, p_2 \)) to point 3 (at \( V_2, p_3 \)), where \( p_3 \) is the new pressure after heat \( Q \) is added.

4. **Label the Diagram**: Clearly label the points as follows:
- Point 1: \( (V_1, p_1) \)
- Point 2: \( (V_2, p_2) \)
- Point 3: \( (V_2, p_3) \)

### Part (ii): Find the total work done by the gas, the total change in its internal energy, and the final temperature of the gas.

#### Step 1: Total Work Done by the Gas

1. **Work Done during Adiabatic Compression (1 to 2)**:
The work done during an adiabatic process is given by:
\[ W_{1 \to 2} = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1} \]
where \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \) for a monoatomic gas.

2. **Finding \( p_2 \)**:
Using the adiabatic condition:
\[ p_1 V_1^\gamma = p_2 V_2^\gamma \]
Rearranging gives:
\[ p_2 = p_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

3. **Substituting \( p_2 \) into Work Done**:
Substitute \( p_2 \) into the work done equation:
\[ W_{1 \to 2} = \frac{p_1 V_1 - p_1 \left( \frac{V_1}{V_2} \right)^\gamma V_2}{\gamma - 1} \]
Simplifying gives:
\[ W_{1 \to 2} = \frac{p_1 V_1}{\gamma - 1} \left( 1 - \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \right) \]

4. **Work Done during Constant Volume Process (2 to 3)**:
Since the volume is constant, the work done \( W_{2 \to 3} = 0 \).

5. **Total Work Done**:
\[ W_{\text{total}} = W_{1 \to 2} + W_{2 \to 3} = W_{1 \to 2} \]

#### Step 2: Total Change in Internal Energy

1. **Change in Internal Energy for Process 1 to 2**:
The change in internal energy during an adiabatic process is given by:
\[ \Delta U_{1 \to 2} = n C_v (T_2 - T_1) \]
For a monoatomic gas, \( C_v = \frac{3}{2} R \).

2. **Change in Internal Energy for Process 2 to 3**:
For the constant volume process, the change in internal energy is equal to the heat added:
\[ \Delta U_{2 \to 3} = Q \]

3. **Total Change in Internal Energy**:
\[ \Delta U_{\text{total}} = \Delta U_{1 \to 2} + \Delta U_{2 \to 3} = n C_v (T_2 - T_1) + Q \]

#### Step 3: Final Temperature of the Gas

1. **Using the Ideal Gas Law**:
The final temperature can be found using the ideal gas law:
\[ pV = nRT \]
For state 1:
\[ T_1 = \frac{p_1 V_1}{nR} \]
For state 2:
\[ T_2 = \frac{p_2 V_2}{nR} \]
For state 3:
\[ T_3 = \frac{p_3 V_2}{nR} \]

2. **Relate \( T_3 \) to \( Q \)**:
Using the relation for change in internal energy and the heat added, we can express \( T_3 \) in terms of \( Q \):
\[ \Delta U_{\text{total}} = n C_v (T_3 - T_1) = Q \]

3. **Final Expression for \( T_3 \)**:
Rearranging gives:
\[ T_3 = T_1 + \frac{Q}{n C_v} \]

### Final Answers

- **Total Work Done**:
\[ W_{\text{total}} = \frac{p_1 V_1}{\gamma - 1} \left( 1 - \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \right) \]

- **Total Change in Internal Energy**:
\[ \Delta U_{\text{total}} = n C_v (T_2 - T_1) + Q \]

- **Final Temperature**:
\[ T_3 = T_1 + \frac{Q}{n C_v} \]