For a process, relation between temperature and volume is `TV^(3)`=constant. If a monoatomic gas follows this process, find molar specific heat for this process

Options

A. `(7R)/(6)`
B. `(R)/(3)`
C. `(11R)/(6)`
D. Zero

Answer

To find the molar specific heat for the given process where the relation between temperature and volume is \( TV^3 = \text{constant} \) for a monoatomic gas, we can follow these steps:

### Step 1: Understand the given relation
We have the relation \( TV^3 = C \), where \( C \) is a constant. This implies that as the volume changes, the temperature will also change in such a way that the product \( TV^3 \) remains constant.

### Step 2: Use the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
From this, we can express temperature \( T \) in terms of pressure \( P \) and volume \( V \):
\[ T = \frac{PV}{nR} \]

### Step 3: Substitute \( T \) into the given relation
Substituting the expression for \( T \) into the relation \( TV^3 = C \):
\[ \left(\frac{PV}{nR}\right)V^3 = C \]
This simplifies to:
\[ \frac{PV^4}{nR} = C \]
Thus, we can write:
\[ PV^4 = nRC \]
Since \( nR \) is a constant, we can denote \( nRC \) as another constant \( C' \):
\[ PV^4 = C' \]

### Step 4: Identify the polytropic index
The equation \( PV^n = \text{constant} \) indicates a polytropic process. By comparing:
\[ PV^4 = C' \]
we see that the polytropic index \( n = 4 \).

### Step 5: Determine the specific heat capacity
For a polytropic process, the molar specific heat capacity \( C \) is given by:
\[ C = \frac{R}{\gamma - 1} + \frac{R}{1 - n} \]
For a monoatomic gas, the ratio of specific heats \( \gamma \) is:
\[ \gamma = \frac{5}{3} \]
Substituting \( n = 4 \) and \( \gamma = \frac{5}{3} \) into the specific heat capacity formula:
\[ C = \frac{R}{\frac{5}{3} - 1} + \frac{R}{1 - 4} \]

### Step 6: Simplify the equation
Calculating the first term:
\[ \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]
Thus:
\[ \frac{R}{\frac{2}{3}} = \frac{3R}{2} \]
Calculating the second term:
\[ 1 - 4 = -3 \quad \Rightarrow \quad \frac{R}{-3} = -\frac{R}{3} \]
Now combining both terms:
\[ C = \frac{3R}{2} - \frac{R}{3} \]
Finding a common denominator (which is 6):
\[ C = \frac{9R}{6} - \frac{2R}{6} = \frac{7R}{6} \]

### Final Answer
The molar specific heat for this process is:
\[ C = \frac{7R}{6} \]