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 equation of process is

Options

A. `1/P=(V_(0)T_(0)^(3//2))/(4RT^(5/2))e^((T-T_(0)/(T_(0)))`
B. `1/P=(V_(0)T_(0)^(3//2))/(2RT^(5/2))e^((T-T_(0)/(T_(0)))`
C. `1/P=(V_(0)T_(0)^(3//2))/(4RT^(5/2))e^((T+T_(0)/(T_(0)))`
D. `1/P=(V_(0)T_(0)^(3//2))/(3RT^(5/2))e^((T-T_(0)/(T_(0)))`

Explanation

To solve the problem, we need to find the equation of the process for the given thermodynamic conditions of the ideal monoatomic gas. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
We have:
- 2 moles of an ideal monoatomic gas.
- The molar heat capacity \( C \) depends on absolute temperature \( T \) as \( C = \frac{RT}{T_0} \), where \( R \) is the gas constant and \( T_0 \) is the initial temperature.
- \( V_0 \) is the initial volume of the gas.

### 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.

### Step 3: Express \( dq \), \( dU \), and \( dW \)
For an ideal gas, we can express:
\[ dq = nC dT \]
\[ dU = nC_v dT \]
\[ dW = P dV \]
Where \( C_v \) for a monoatomic gas is \( \frac{3}{2}R \).

### Step 4: Substitute the Values
Substituting \( n = 2 \) moles and \( C \):
\[ dq = 2 \cdot \frac{RT}{T_0} dT \]
\[ dU = 2 \cdot \frac{3}{2}R dT = 3R dT \]
Using the ideal gas law \( PV = nRT \), we can express pressure \( P \):
\[ P = \frac{nRT}{V} = \frac{2RT}{V} \]
Thus, the work done becomes:
\[ dW = \frac{2RT}{V} dV \]

### Step 5: Combine the Equations
Now substituting these into the first law equation:
\[ 2 \cdot \frac{RT}{T_0} dT = 3R dT + \frac{2RT}{V} dV \]
Dividing through by \( R \):
\[ \frac{2T}{T_0} dT = 3 dT + \frac{2T}{V} dV \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \left(\frac{2}{T_0} - 3\right) dT = \frac{2}{V} dV \]
This can be simplified to:
\[ \frac{1}{T_0 - \frac{3}{2}T} dT = \frac{1}{V} dV \]

### Step 7: Integrate Both Sides
Integrating both sides with limits \( T_0 \) to \( T \) and \( V_0 \) to \( V \):
\[ \int_{T_0}^{T} \frac{1}{T_0 - \frac{3}{2}T} dT = \int_{V_0}^{V} \frac{1}{V} dV \]
This results in:
\[ \ln\left|T_0 - \frac{3}{2}T\right| = \ln\left|V\right| - \ln\left|V_0\right| + C \]

### Step 8: Solve for the Equation of the Process
Exponentiating both sides gives:
\[ T_0 - \frac{3}{2}T = \frac{V}{V_0} e^C \]
Rearranging gives the equation of the process.

### Final Equation
The final equation can be expressed in terms of \( V \) and \( T \), leading to:
\[ V = V_0 \left(\frac{T_0}{T}\right)^{\frac{3}{2}} e^{\frac{T - T_0}{T_0}} \]