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 process where the relation between temperature (T) and volume (V) is given by \( TV^3 = \text{constant} \), we will follow these steps:

### Step 1: Express the relationship mathematically
Given the relationship \( TV^3 = C \) (where C is a constant), we can express temperature in terms of volume:
\[ T = \frac{C}{V^3} \]

### Step 2: Use the first law of thermodynamics
The first law of thermodynamics states:
\[ dQ = dU + dW \]
where \( dQ \) is the heat added, \( dU \) is the change in internal energy, and \( dW \) is the work done by the system.

### Step 3: Calculate the change in internal energy (\( dU \))
For a monoatomic ideal gas, the change in internal energy is given by:
\[ dU = nC_V dT \]
where \( C_V = \frac{3}{2} R \) for a monoatomic gas.

### Step 4: Calculate the work done (\( dW \))
The work done by the gas during an infinitesimal expansion is given by:
\[ dW = PdV \]
Using the ideal gas law, \( PV = nRT \), we can substitute for \( P \):
\[ P = \frac{nRT}{V} \]
Substituting \( T \) from our earlier expression:
\[ P = \frac{nR \left( \frac{C}{V^3} \right)}{V} = \frac{nRC}{V^4} \]
Thus, the work done becomes:
\[ dW = \frac{nRC}{V^4} dV \]

### Step 5: Substitute \( dU \) and \( dW \) into the first law
Now substituting \( dU \) and \( dW \) into the first law:
\[ dQ = nC_V dT + dW \]
Substituting for \( dW \):
\[ dQ = nC_V dT + \frac{nRC}{V^4} dV \]

### Step 6: Find \( dT \) in terms of \( dV \)
From \( T = \frac{C}{V^3} \), we differentiate:
\[ dT = -\frac{3C}{V^4} dV \]

### Step 7: Substitute \( dT \) into the heat equation
Substituting \( dT \) into the equation for \( dQ \):
\[ dQ = nC_V \left(-\frac{3C}{V^4} dV\right) + \frac{nRC}{V^4} dV \]
\[ dQ = \left(-3nC_V + nR\right) \frac{C}{V^4} dV \]

### Step 8: Identify the molar specific heat (\( C \))
The molar specific heat \( C \) for the process can be defined as:
\[ C = \frac{dQ}{dT} \]
Using the relationship from above, we can express \( C \) in terms of \( n \):
\[ C = -3C_V + R \]
Substituting \( C_V = \frac{3}{2}R \):
\[ C = -3 \left(\frac{3}{2}R\right) + R = -\frac{9}{2}R + R = -\frac{7}{2}R \]

### Step 9: Final result
Thus, the molar specific heat for the process is:
\[ C = -\frac{7}{2} R \]