A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats `gamma`. It is moving with speed v and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:

Options

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

Explanation

To solve the problem, we need to analyze the situation step by step using the principles of thermodynamics and kinetic energy.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- We have an ideal gas in a thermally insulated vessel.
- The gas has a molecular mass \( M \) and a specific heat ratio \( \gamma \).
- The vessel is moving with speed \( v \).

2. **Calculate the Initial Kinetic Energy**:
- The kinetic energy (KE) of the moving vessel is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- Here, \( m \) is the mass of the gas in the vessel.

3. **Final Conditions After Stopping**:
- When the vessel is suddenly brought to rest, its final kinetic energy becomes zero.
- Since the vessel is thermally insulated, no heat is lost to the surroundings.

4. **Apply the First Law of Thermodynamics**:
- According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
- Since the vessel is insulated, \( Q = 0 \). Thus:
\[ 0 = \Delta U + W \]
- Rearranging gives:
\[ \Delta U = -W \]

5. **Relate Work Done to Kinetic Energy Change**:
- The work done on the gas is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_{final} - KE_{initial} = 0 - \frac{1}{2} mv^2 = -\frac{1}{2} mv^2 \]
- Therefore:
\[ \Delta U = \frac{1}{2} mv^2 \]

6. **Relate Change in Internal Energy to Temperature Change**:
- The change in internal energy \( \Delta U \) for an ideal gas can also be expressed in terms of temperature change:
\[ \Delta U = n C_v \Delta T \]
- Where \( n \) is the number of moles and \( C_v \) is the molar specific heat at constant volume. For an ideal gas:
\[ C_v = \frac{R}{\gamma - 1} \]
- Here, \( n = \frac{m}{M} \) (where \( M \) is the molecular mass).

7. **Substituting Values**:
- Substitute \( n \) and \( C_v \) into the equation:
\[ \frac{1}{2} mv^2 = \left(\frac{m}{M}\right) \left(\frac{R}{\gamma - 1}\right) \Delta T \]

8. **Solving for Temperature Change \( \Delta T \)**:
- Rearranging gives:
\[ \Delta T = \frac{\frac{1}{2} mv^2 (\gamma - 1)}{\frac{mR}{M}} = \frac{Mv^2 (\gamma - 1)}{2R} \]

9. **Final Result**:
- Thus, the increase in temperature of the gas is:
\[ \Delta T = \frac{Mv^2 (\gamma - 1)}{2R} \]