A quantity of 2 mole of helium gas undergoes a thermodynamic process, in which molar specific heat capacity of the gas depends on absolute temperature` T` , according to relation:
`C=(3RT)/(4T_(0)`
where `T_(0)` is initial temperature of gas. It is observed that when temperature is increased. volume of gas first decrease then increase. The total work done on the gas until it reaches minimum volume is :-

Options

A. `(3)/(2) RT_(0)`
B. `(3)/(4)RT_(0)`
C. `(3)/(8)RT_(0)`
D. `(3)/(10)RT_(0)`

Explanation

To find the total work done on the gas until it reaches its minimum volume, we can follow these steps:

### Step 1: Understand the Given Information
We have 2 moles of helium gas, and the molar specific heat capacity \( C \) is given by the relation:
\[ C = \frac{3RT}{4T_0} \]
where \( T_0 \) is the initial temperature of the gas.

### Step 2: Identify the Process
It is observed that when the temperature increases, the volume of the gas first decreases and then increases. This indicates that there is a minimum volume at a certain temperature. At this minimum volume, the change in volume with respect to temperature is zero:
\[ \frac{dV}{dT} = 0 \]
This implies that the process is isochoric (constant volume) at the minimum point.

### Step 3: Determine the Final Temperature
For an isochoric process, the specific heat capacity \( C_v \) for a monatomic gas like helium is given by:
\[ C_v = \frac{3}{2} R \]
We can equate the two expressions for specific heat capacity at the minimum volume:
\[ \frac{3RT}{4T_0} = \frac{3}{2} R \]
Cancelling \( R \) from both sides, we get:
\[ \frac{3T}{4T_0} = \frac{3}{2} \]
Solving for \( T \):
\[ T = 2T_0 \]

### Step 4: Calculate the Heat Added to the Gas
The heat added to the gas can be calculated using the formula:
\[ Q = nC \Delta T \]
Substituting the values:
\[ Q = 2 \cdot \frac{3R}{4T_0} \cdot (2T_0 - T_0) \]
\[ Q = 2 \cdot \frac{3R}{4T_0} \cdot T_0 = \frac{3RT_0}{2} \]

### Step 5: Calculate the Change in Internal Energy
The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = nC_v \Delta T \]
Substituting the values:
\[ \Delta U = 2 \cdot \frac{3}{2} R \cdot (2T_0 - T_0) = 2 \cdot \frac{3}{2} R \cdot T_0 = 3RT_0 \]

### Step 6: Calculate the Work Done on the Gas
The work done on the gas can be calculated using the first law of thermodynamics:
\[ W = Q - \Delta U \]
Substituting the values we found:
\[ W = \frac{3RT_0}{2} - 3RT_0 \]
\[ W = \frac{3RT_0}{2} - \frac{6RT_0}{2} = -\frac{3RT_0}{2} \]

### Conclusion
The total work done on the gas until it reaches minimum volume is:
\[ W = -\frac{3RT_0}{2} \]
The negative sign indicates that work is done on the gas.