Two moles of an ideal mono-atomic gas undergoes a thermodynamic process in which the molar heat capacity `'C'` of the gas depends on absolute temperature as `C=(RT)/(T_(0))`, where `R` is gas consant and `T_(0)` is the initial temperature of the gas. (`V_(0)` is the initial of the gas). Then answer the following questions:
The minimum volume of gas is

Options

A. `(2/3)^(3//2)V_(0)e^(1//2)`
B. `(4/3)^(3//2)V_(0)e^(1//2)`
C. `(5/3)^(3//2)V_(0)e^(1//2)`
D. `(8/3)^(3//2)V_(0)e^(1//2)`

Explanation

To find the minimum volume of the gas undergoing a thermodynamic process, we can follow these steps:

### Step 1: Understanding the Problem
We are given that the molar heat capacity \( C \) of the gas depends on the absolute temperature \( T \) as:
\[ C = \frac{RT}{T_0} \]
where \( R \) is the gas constant, \( T_0 \) is the initial temperature, and the gas is monoatomic with 2 moles.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ dQ = dU + dW \]
where:
- \( dQ \) is the change in heat,
- \( dU \) is the change in internal energy,
- \( dW \) is the work done.

For an ideal gas, we can express \( dQ \) as:
\[ dQ = nC dT \]
and the change in internal energy \( dU \) for a monoatomic ideal gas is given by:
\[ dU = nC_v dT = n \left(\frac{3}{2} R\right) dT \]
where \( C_v \) is the molar heat capacity at constant volume.

### Step 3: Express Work Done
The work done \( dW \) can be expressed as:
\[ dW = P dV \]
For an ideal gas, we know from the ideal gas law that:
\[ PV = nRT \implies P = \frac{nRT}{V} \]
Substituting this into the equation for work done, we have:
\[ dW = \frac{nRT}{V} dV \]

### Step 4: Substitute Values into the First Law
Substituting \( dQ \), \( dU \), and \( dW \) into the first law equation gives:
\[ nC dT = nC_v dT + \frac{nRT}{V} dV \]
Substituting \( C = \frac{RT}{T_0} \) and \( C_v = \frac{3}{2} R \):
\[ n \left(\frac{RT}{T_0}\right) dT = n \left(\frac{3}{2} R\right) dT + \frac{nRT}{V} dV \]

### Step 5: Rearranging the Equation
Dividing through by \( n \) and rearranging gives:
\[ \left(\frac{RT}{T_0} - \frac{3}{2} R\right) dT = \frac{RT}{V} dV \]
Simplifying further, we have:
\[ \left(\frac{R}{T_0}T - \frac{3R}{2}\right) dT = \frac{RT}{V} dV \]
Cancelling \( R \) from both sides (assuming \( R \neq 0 \)):
\[ \left(\frac{T}{T_0} - \frac{3}{2}\right) dT = \frac{T}{V} dV \]

### Step 6: Separate Variables and Integrate
Rearranging gives:
\[ \frac{dV}{V} = \frac{dT}{\left(\frac{T}{T_0} - \frac{3}{2}\right)} \]
Integrating both sides with appropriate limits will yield the relationship between \( V \) and \( T \).

### Step 7: Finding Minimum Volume
To find the minimum volume, differentiate the volume expression with respect to temperature \( T \) and set the derivative equal to zero:
\[ \frac{dV}{dT} = 0 \]
This will give us the temperature \( T \) at which the volume is minimized.

### Step 8: Substitute Back to Find Minimum Volume
After finding the critical temperature, substitute it back into the volume equation to find the minimum volume \( V_{min} \).

### Final Expression for Minimum Volume
The final expression for the minimum volume of the gas is:
\[ V_{min} = V_0 \left( \frac{2}{3} \right)^{3/2} e^{\frac{1}{2}} \]