A monatomic gas undergoes a thermodynamic process such that `Tprop V^(-2/7)` (T = temperature and V = volume of the gas) As heat is supplied to the gas, choose INCORRECT option.

Options

A. ms velocity of gas molecules increases
B. rms velocity of gas molecules decreases
C. work done on the gas is positive
D. density of the gas increases

Answer

To solve the problem, we start by analyzing the given relationship between temperature (T) and volume (V) for a monatomic gas undergoing a thermodynamic process, where \( T \propto V^{-2/7} \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
Given that \( T \propto V^{-2/7} \), we can express this as:
\[ T = k \cdot V^{-2/7} \]
where \( k \) is a constant.

2. **Heat Supplied**:
Since heat is being supplied to the gas, we know that the heat transfer \( Q > 0 \). This implies that the temperature of the gas must increase.

3. **RMS Velocity**:
The root mean square (RMS) velocity of the gas molecules is given by:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant and \( M \) is the molar mass of the gas. Since \( R \) and \( M \) are constants, \( V_{\text{rms}} \) is directly proportional to the square root of the temperature \( T \).

4. **Effect of Temperature Increase**:
As heat is supplied and the temperature \( T \) increases, the RMS velocity \( V_{\text{rms}} \) will also increase. This means that any option stating that the RMS velocity decreases is incorrect.

5. **Analyzing Work Done**:
For a polytropic process where \( PV^x = \text{constant} \), we can relate the work done on the gas. Here, we find \( x \) from the relationship:
\[ PV^{9/7} = \text{constant} \]
The work done on the gas can be expressed as:
\[ W = \frac{nR\Delta T}{1 - x} \]
Substituting \( x = \frac{9}{7} \):
\[ W = \frac{nR\Delta T}{1 - \frac{9}{7}} = \frac{nR\Delta T}{-\frac{2}{7}} \quad (\text{which is negative}) \]
Thus, the work done by the gas is negative, indicating that work is done on the gas (positive work done on the gas).

6. **Density Change**:
The density \( \rho \) of the gas is given by:
\[ \rho = \frac{m}{V} \]
As the temperature increases, from the relationship \( T \propto V^{-2/7} \), we can infer that the volume \( V \) must decrease. A decrease in volume while mass remains constant leads to an increase in density.

7. **Conclusion**:
Based on the analysis, the incorrect option is the one stating that the RMS velocity of gas molecules decreases, as we have established that it actually increases with an increase in temperature.

### Final Answer:
The incorrect option is: **RMS Velocity of Gas Molecule Decreases**.