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

Answer

To solve the problem, we need to analyze the situation where an ideal gas is contained in a closed vessel that is moving with a constant velocity \( v \). When the vessel is suddenly stopped, the kinetic energy of the gas will convert into thermal energy, causing a rise in temperature.

### Step-by-Step Solution:

1. **Identify the initial kinetic energy of the gas:**
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 total mass of the gas.

2. **Understand the conversion of kinetic energy to thermal energy:**
When the vessel is suddenly stopped, the kinetic energy of the gas is converted into internal energy, which raises the temperature of the gas. This process can be considered adiabatic since there is no heat exchange with the surroundings.

3. **Relate the change in internal energy to temperature change:**
For an ideal gas, the change in internal energy \( \Delta U \) is related to the change in temperature \( \Delta T \) by the equation:
\[ \Delta U = n C_V \Delta T \]
where \( n \) is the number of moles of gas, \( C_V \) is the molar specific heat at constant volume, and \( \Delta T \) is the change in temperature.

4. **Express the number of moles in terms of mass and molar mass:**
The number of moles \( n \) can be expressed as:
\[ n = \frac{m}{M} \]
where \( M \) is the molecular mass of the gas.

5. **Substitute the expression for \( n \) into the internal energy equation:**
Substituting \( n \) into the internal energy equation gives:
\[ \Delta U = \frac{m}{M} C_V \Delta T \]

6. **Set the kinetic energy equal to the change in internal energy:**
Since the kinetic energy is converted into internal energy, we have:
\[ \frac{1}{2} mv^2 = \frac{m}{M} C_V \Delta T \]

7. **Cancel the mass \( m \) from both sides:**
\[ \frac{1}{2} v^2 = \frac{C_V}{M} \Delta T \]

8. **Solve for the change in temperature \( \Delta T \):**
Rearranging the equation gives:
\[ \Delta T = \frac{Mv^2}{2C_V} \]

9. **Relate \( C_V \) to \( C_P \) using the relation \( \gamma = \frac{C_P}{C_V} \):**
We know that:
\[ C_P = \gamma C_V \]
Thus, we can express \( C_V \) in terms of \( C_P \):
\[ C_V = \frac{C_P}{\gamma} \]

10. **Substitute \( C_V \) back into the temperature change equation:**
\[ \Delta T = \frac{Mv^2}{2 \left(\frac{C_P}{\gamma}\right)} = \frac{M \gamma v^2}{2 C_P} \]

11. **Final expression for the rise in temperature:**
The rise in temperature of the gas when the vessel is suddenly stopped is:
\[ \Delta T = \frac{\gamma M v^2}{2 C_P} \]

### Conclusion:
The rise in temperature of the gas when the vessel is suddenly stopped is given as:
\[ \Delta T = \frac{\gamma v^2}{2 R} \quad \text{(where \( R \) is the gas constant)} \]