An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy. If in the process, the initial temperature of the gas be `T_(0)` and the final volume by 32 times the initial volume, the work done `(` in Joules `)` by the gas during the process will be

Options

A. `RT_(0)`
B. `(5RT_(0))/(2)`
C. `2RT_(0)`
D. `(RT_(0))/(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between heat, internal energy, and work done
Given that the heat transferred to the gas (dQ) is equal to the decrease in its internal energy (dU), we can write:
\[ dQ = -dU \]

From the first law of thermodynamics, we know:
\[ dQ = dU + dW \]
Substituting \( dQ \) from the first equation into this gives:
\[ -dU = dU + dW \]
This simplifies to:
\[ dW = 2dU \]

### Step 2: Express dU in terms of temperature change
For an ideal gas, the change in internal energy (dU) can be expressed as:
\[ dU = nC_v dT \]
where \( C_v \) is the specific heat at constant volume.

### Step 3: Relate work done to temperature change
Since we have \( dW = 2dU \), we can write:
\[ dW = 2(nC_v dT) \]

### Step 4: Use the ideal gas law to express pressure
From the ideal gas law, we know:
\[ PV = nRT \]
Thus, we can express pressure (P) as:
\[ P = \frac{nRT}{V} \]

### Step 5: Relate volume change to temperature change
We know that the final volume \( V_2 \) is 32 times the initial volume \( V_1 \):
\[ V_2 = 32V_1 \]

Using the relation derived earlier, we can integrate the equations to find the relationship between the initial and final temperatures and volumes:
\[ \frac{V_1}{V_2}^{\frac{\gamma - 1}{2}} = \frac{T_2}{T_1} \]

### Step 6: Substitute the known values
For a diatomic gas, \( \gamma = \frac{7}{5} \). Substituting \( V_2 = 32V_1 \) into the equation gives:
\[ \left(\frac{1}{32}\right)^{\frac{7/5 - 1}{2}} = \frac{T_2}{T_0} \]

### Step 7: Solve for T2
After calculating, we find:
\[ T_2 = T_0 \left(\frac{1}{32}\right)^{\frac{1}{5}} = \frac{T_0}{2} \]

### Step 8: Calculate the work done
Now substituting \( T_2 \) back into the work done equation:
\[ dW = 2nC_v(T_2 - T_0) \]
Substituting \( C_v = \frac{R}{\gamma - 1} \) and using \( n = 1 \):
\[ dW = 2 \cdot \frac{R}{\frac{7}{5} - 1} \cdot \left(\frac{T_0}{2} - T_0\right) \]
This simplifies to:
\[ dW = 2 \cdot \frac{5R}{2} \cdot \left(-\frac{T_0}{2}\right) \]
\[ dW = 5RT_0/2 \]

### Final Answer
Thus, the work done by the gas during the process is:
\[ \boxed{\frac{5RT_0}{2}} \]