A certain amount of ideal monoatomic gas undergoes a process given by `TV^(1//2)`= constant. The molar specific heat of the gas for the process will be given by

Options

A. R/2
B. `-3R//2`
C. `3R//2`
D. `-R//2`

Answer

To solve the problem, we need to find the molar specific heat of an ideal monoatomic gas undergoing a process defined by the equation \( TV^{1/2} = \text{constant} \).

### Step-by-step Solution:

1. **Understanding the Process**:
We start with the equation given in the problem:
\[ TV^{1/2} = K \quad (1) \]
where \( K \) is a constant.

2. **Relating Temperature and Pressure**:
We know from the ideal gas law that:
\[ PV = nRT \quad (2) \]
Rearranging this gives:
\[ T = \frac{PV}{nR} \quad (3) \]

3. **Substituting Temperature into the Process Equation**:
Substitute equation (3) into equation (1):
\[ \left(\frac{PV}{nR}\right)V^{1/2} = K \]
This simplifies to:
\[ \frac{PV^{3/2}}{nR} = K \]
Rearranging gives:
\[ PV^{3/2} = nRK \quad (4) \]
This shows that \( PV^{3/2} \) is also constant.

4. **Identifying the Value of Gamma**:
The equation \( PV^{\gamma} = \text{constant} \) allows us to identify \( \gamma \):
\[ \gamma = \frac{3}{2} \quad (5) \]

5. **Using the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ \delta Q = \delta U + \delta W \quad (6) \]
We know that:
\[ \delta Q = nC \Delta T \quad (7) \]
where \( C \) is the molar specific heat we want to find.

6. **Calculating Change in Internal Energy**:
For a monoatomic gas, the change in internal energy is given by:
\[ \delta U = nC_V \Delta T \quad (8) \]
where \( C_V = \frac{3}{2}R \) for a monoatomic gas.

7. **Calculating Work Done**:
The work done can be expressed as:
\[ \delta W = nR \Delta T \cdot \frac{1}{1 - \gamma} \quad (9) \]
Substituting \( \gamma = \frac{3}{2} \):
\[ \delta W = nR \Delta T \cdot \frac{1}{1 - \frac{3}{2}} = nR \Delta T \cdot \frac{1}{-\frac{1}{2}} = -2nR \Delta T \]

8. **Substituting into the First Law**:
Substitute equations (7), (8), and (9) into equation (6):
\[ nC \Delta T = nC_V \Delta T - 2nR \Delta T \]
Dividing through by \( n \Delta T \) (assuming \( \Delta T \neq 0 \)):
\[ C = C_V - 2R \quad (10) \]

9. **Substituting \( C_V \)**:
Now substitute \( C_V = \frac{3}{2}R \) into equation (10):
\[ C = \frac{3}{2}R - 2R = \frac{3}{2}R - \frac{4}{2}R = -\frac{1}{2}R \]

10. **Final Result**:
Thus, the molar specific heat \( C \) for the process is:
\[ C = -\frac{R}{2} \]

### Conclusion:
The molar specific heat of the gas for the process is \( -\frac{R}{2} \).