One mole of an ideal gas is warmed slowly so that it goes form the `PV` state (`P_(i) V_(i)`) to `(3P_(i), 3V_(i)`) in such a way that the pressure of the gas is directly proportional to the volume.
(a) How much work is done on the gas in the process?
(b) How is the temperature of the gas related to its volume during this process?

Answer

To solve the problem, we will break it down into two parts as stated in the question.

### Part (a): Work Done on the Gas

1. **Identify the Initial and Final States**:
- Initial state: \( P_i, V_i \)
- Final state: \( 3P_i, 3V_i \)

2. **Understand the Relationship**:
- The pressure \( P \) is directly proportional to the volume \( V \). This can be expressed as:
\[ P = kV \]
- From the initial state, we can find \( k \):
\[ P_i = kV_i \implies k = \frac{P_i}{V_i} \]

3. **Equation of the Line**:
- The equation of the line in the \( PV \) diagram is:
\[ P = \frac{P_i}{V_i} V \]

4. **Calculate Work Done**:
- The work done on the gas during the process can be calculated as the negative of the area under the \( PV \) curve.
- The area under the curve forms a trapezium with bases \( P_i \) and \( 3P_i \) and height \( (3V_i - V_i) = 2V_i \).
- The area \( A \) of the trapezium is given by:
\[ A = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \]
\[ A = \frac{1}{2} \times (P_i + 3P_i) \times 2V_i = \frac{1}{2} \times 4P_i \times 2V_i = 4P_i V_i \]
- Therefore, the work done on the gas is:
\[ W = -A = -4P_i V_i \]

### Part (b): Relationship Between Temperature and Volume

1. **Using Ideal Gas Law**:
- The ideal gas law states:
\[ PV = nRT \]
- For one mole of gas (\( n = 1 \)):
\[ PV = RT \]

2. **Substituting for Pressure**:
- From the relationship \( P = \frac{P_i}{V_i} V \):
\[ \left(\frac{P_i}{V_i} V\right) V = RT \]
- Simplifying gives:
\[ \frac{P_i}{V_i} V^2 = RT \]

3. **Expressing Temperature**:
- Rearranging the equation gives:
\[ T = \frac{P_i}{R V_i} V^2 \]
- This shows that temperature \( T \) is proportional to the square of the volume \( V \):
\[ T \propto V^2 \]

### Final Answers:
(a) The work done on the gas is \( W = -4P_i V_i \).

(b) The temperature of the gas is related to its volume by \( T \propto V^2 \).