The relation between the internal energy (U) versus temeperature (T) for a 3 mole ideal gas in an isochoric process is given as `U=alpha+betaT` where `alpha` and `beta` are constant. If the specific heat capacity of the gas for above process is `(beta)/(m)` find the value of m.

Explanation

To solve the problem, we need to analyze the relationship between the internal energy (U) and temperature (T) for a 3-mole ideal gas in an isochoric process, where the internal energy is given by the equation:

\[ U = \alpha + \beta T \]

Here, \(\alpha\) and \(\beta\) are constants.

### Step 1: Understand the Isochoric Process
In an isochoric process, the volume of the gas remains constant. This means that no work is done on or by the gas, as work done (W) is given by:

\[ W = P \Delta V \]

Since \(\Delta V = 0\), we have \(W = 0\).

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:

\[ \Delta U = Q - W \]

Since \(W = 0\) in an isochoric process, we have:

\[ \Delta U = Q \]

This means that the change in internal energy is equal to the heat added to the system.

### Step 3: Differentiate the Internal Energy Equation
To find the specific heat capacity, we need to differentiate the internal energy equation with respect to temperature (T):

\[ \frac{dU}{dT} = \frac{d}{dT}(\alpha + \beta T) = \beta \]

### Step 4: Define Specific Heat Capacity
The specific heat capacity at constant volume (\(C_V\)) is defined as the amount of heat required to change the temperature of the gas by one degree per mole. Mathematically, it can be expressed as:

\[ C_V = \frac{Q}{n \Delta T} \]

For an isochoric process, we can express the heat added as:

\[ Q = n \cdot C_V \cdot \Delta T \]

### Step 5: Relate Heat to Internal Energy Change
From our previous step, we know that:

\[ Q = \Delta U \]

Thus, we can write:

\[ \Delta U = n \cdot C_V \cdot \Delta T \]

Substituting \(\Delta U\) with \(\beta \Delta T\):

\[ \beta \Delta T = n \cdot C_V \cdot \Delta T \]

### Step 6: Cancel \(\Delta T\) and Solve for \(C_V\)
Assuming \(\Delta T \neq 0\), we can cancel \(\Delta T\) from both sides:

\[ \beta = n \cdot C_V \]

### Step 7: Substitute for \(C_V\)
We are given that the specific heat capacity for the gas is:

\[ C_V = \frac{\beta}{m} \]

Substituting this into our equation gives:

\[ \beta = n \cdot \frac{\beta}{m} \]

### Step 8: Solve for \(m\)
Rearranging the equation, we have:

\[ m = n \]

Given that \(n = 3\) moles (as per the problem statement), we find:

\[ m = 3 \]

### Final Answer
Thus, the value of \(m\) is:

\[ \boxed{3} \]