An ideal diatomic gas with `C_V = (5 R)/2` occupies a volume `(V_(i)` at a pressure `(P_(i)`. The gas undergoes a process in which the pressure is proportional to the volume. At the end of the process, it is found that the rms speed of the gas molecules has doubles from its initial value. Determine the amount of energy transferred to the gas by heat.

Explanation

To solve the problem, we need to analyze the given information step by step.

### Step 1: Understand the Initial Conditions
We have an ideal diatomic gas with the following properties:
- Heat capacity at constant volume, \( C_V = \frac{5R}{2} \)
- Initial volume, \( V_i \)
- Initial pressure, \( P_i \)

### Step 2: Relationship Between Pressure and Volume
The problem states that pressure is proportional to volume:
\[ P \propto V \implies P = kV \]
for some constant \( k \). This implies that \( PV = kV^2 \), which means that the process follows the relation \( PV^x = \text{constant} \) with \( x = -1 \).

### Step 3: Change in RMS Speed
The problem states that the root mean square (RMS) speed of the gas molecules doubles:
\[ v_{\text{rms, final}} = 2 v_{\text{rms, initial}} \]
The RMS speed for an ideal gas is given by:
\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \quad \text{(for monatomic gas)} \]
For diatomic gas, it can be expressed as:
\[ v_{\text{rms}} = \sqrt{\frac{5RT}{M}} \]
Since the RMS speed doubles, we have:
\[ \sqrt{\frac{5R(T_f)}{M}} = 2 \sqrt{\frac{5R(T_i)}{M}} \]
Squaring both sides gives:
\[ \frac{5R(T_f)}{M} = 4 \cdot \frac{5R(T_i)}{M} \]
This simplifies to:
\[ T_f = 4T_i \]

### Step 4: Determine the Change in Temperature
The change in temperature is:
\[ \Delta T = T_f - T_i = 4T_i - T_i = 3T_i \]

### Step 5: Calculate the Heat Transfer
The amount of heat transferred to the gas can be calculated using the formula:
\[ Q = N C \Delta T \]
Where:
- \( N \) is the number of moles of gas
- \( C \) is the effective heat capacity for the process

Since we found that \( C \) for the process is:
\[ C = C_V + \frac{R}{1 - x} = \frac{5R}{2} + \frac{R}{1 - (-1)} = \frac{5R}{2} + \frac{R}{2} = 3R \]
Thus, substituting the values:
\[ Q = N \cdot 3R \cdot (3T_i) = 9NRT_i \]

### Final Answer
The amount of energy transferred to the gas by heat is:
\[ Q = 9NRT_i \]