Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velcity `v`. The molecular mass of gas is `M`. The rise in temperature of the gas when the vessel is suddenly stopped is `(gamma C_(P) // C_(V))`

Options

A. `(Mv^(2) (gamma -1))/(2 R (gamma + 1))`
B. `(Mv^(2) (gamma -1))/(2 R)`
C. `(Mv^(2))/(2 R (gamma + 1))`
D. `(Mv^(2))/(2 R (gamma - 1))`

Explanation

To solve the problem, we need to analyze the situation where a closed vessel containing an ideal gas is moving with a constant velocity \( v \) and then suddenly comes to a stop. The kinetic energy of the gas will be converted into internal energy, leading to a rise in temperature.

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy**:
The kinetic energy \( KE \) of the gas when the vessel is moving with velocity \( v \) can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the gas.

2. **Relate Mass to Molecular Mass**:
The mass \( m \) can be expressed in terms of the amount of gas \( W \) (mass of the gas) and the molecular mass \( M \):
\[ m = \frac{W}{M} \]
Therefore, substituting this into the kinetic energy equation gives:
\[ KE = \frac{1}{2} \left(\frac{W}{M}\right) v^2 \]

3. **Convert Kinetic Energy to Internal Energy**:
When the vessel stops, this kinetic energy is converted into internal energy, which can be described by the change in internal energy \( \Delta U \):
\[ \Delta U = n C_V \Delta T \]
where \( n \) is the number of moles of gas, \( C_V \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature.

4. **Express Number of Moles**:
The number of moles \( n \) can be expressed as:
\[ n = \frac{W}{M} \]
Substituting this into the internal energy equation gives:
\[ \Delta U = \frac{W}{M} C_V \Delta T \]

5. **Set Kinetic Energy Equal to Change in Internal Energy**:
Equating the kinetic energy and the change in internal energy:
\[ \frac{1}{2} \left(\frac{W}{M}\right) v^2 = \frac{W}{M} C_V \Delta T \]

6. **Cancel Out Common Terms**:
Since \( W \) and \( M \) are common on both sides, we can cancel them out:
\[ \frac{1}{2} v^2 = C_V \Delta T \]

7. **Solve for Change in Temperature**:
Rearranging the equation gives:
\[ \Delta T = \frac{v^2}{2 C_V} \]

8. **Relate \( C_P \) and \( C_V \) Using \( \gamma \)**:
We know that \( \gamma = \frac{C_P}{C_V} \) and also that:
\[ C_P - C_V = R \]
From this, we can express \( C_P \) in terms of \( C_V \):
\[ C_P = \gamma C_V \]

9. **Substitute \( C_V \) in Terms of \( R \) and \( \gamma \)**:
Rearranging gives:
\[ C_V = \frac{R}{\gamma - 1} \]

10. **Substituting Back to Find \( \Delta T \)**:
Substituting \( C_V \) back into the equation for \( \Delta T \):
\[ \Delta T = \frac{v^2}{2 \left(\frac{R}{\gamma - 1}\right)} = \frac{v^2 (\gamma - 1)}{2R} \]

### Final Result:
The rise in temperature of the gas when the vessel is suddenly stopped is:
\[ \Delta T = \frac{v^2 (\gamma - 1)}{2R} \]