The molar specific heat of the process ` V alpha T^(4)` for `CH_(4)` gas at room temperature is :-

Options

A. `4R`
B. `7R`
C. `3R`
D. `8R`

Explanation

To find the molar specific heat of the process where the volume \( V \) is proportional to \( T^4 \) for methane gas (\( CH_4 \)), we can follow these steps:

### Step 1: Understand the relationship between \( V \) and \( T \)
Given that \( V \) is proportional to \( T^4 \), we can express this mathematically as:
\[ V = kT^4 \]
where \( k \) is a constant.

### Step 2: Use the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature.

### Step 3: Substitute \( V \) in the ideal gas law
Substituting \( V = kT^4 \) into the ideal gas law gives:
\[ P(kT^4) = nRT \]
This simplifies to:
\[ PkT^4 = nRT \]
From this, we can express \( P \) as:
\[ P = \frac{nR}{k} \cdot \frac{1}{T^3} \]

### Step 4: Identify the specific heat capacities
The molar specific heat \( C \) can be expressed in terms of the specific heat at constant volume \( C_v \) and the specific heat ratio \( \gamma \):
\[ C = C_v + R \cdot \left(1 - \frac{1}{\gamma}\right) \]
For methane (\( CH_4 \)), the specific heat at constant volume \( C_v \) is given as:
\[ C_v = 3R \]

### Step 5: Determine the value of \( \gamma \)
From the relationship \( PV^\gamma = \text{constant} \), we can derive \( \gamma \) using the expression for \( V \):
Since \( V \propto T^4 \), we can relate the pressures and volumes:
\[ PV^3 \propto T^4 \implies P \propto T^3 \]
Thus, we can conclude:
\[ \gamma = \frac{C_p}{C_v} = \frac{4}{3} \]

### Step 6: Substitute \( C_v \) and \( \gamma \) into the specific heat equation
Now substituting \( C_v = 3R \) and \( \gamma = \frac{4}{3} \) into the specific heat equation:
\[ C = 3R + R \cdot \left(1 - \frac{3}{4}\right) \]
This simplifies to:
\[ C = 3R + R \cdot \left(\frac{1}{4}\right) = 3R + \frac{R}{4} = 3R + 0.25R = 3.25R \]

### Final Answer
Thus, the molar specific heat of the process \( V \propto T^4 \) for \( CH_4 \) gas at room temperature is:
\[ C = 7R \]