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

Explanation

To solve the problem, we need to analyze the situation using the principles of thermodynamics. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a thermally insulated vessel containing an ideal gas. The vessel is moving with a speed \( v \) and is suddenly brought to rest. Since the vessel is insulated, no heat is exchanged with the surroundings.

### Step 2: Apply the First Law of Thermodynamics
The First Law of Thermodynamics states:
\[ \Delta Q = \Delta U + W \]
Where:
- \( \Delta Q \) is the heat added to the system (which is zero since the vessel is insulated),
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the system.

Since the vessel is insulated, \( \Delta Q = 0 \). Also, because the volume does not change when the vessel is brought to rest, the work done \( W = 0 \).

Thus, we have:
\[ 0 = \Delta U + 0 \implies \Delta U = 0 \]

### Step 3: Relate Kinetic Energy to Internal Energy
When the vessel is brought to rest, the kinetic energy of the gas molecules is converted into internal energy. The kinetic energy \( KE \) of the gas can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of the gas.

### Step 4: Express Change in Internal Energy
The change in internal energy \( \Delta U \) for an ideal gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
Where:
- \( n \) is the number of moles of the gas,
- \( C_v \) is the molar heat capacity at constant volume,
- \( \Delta T \) is the change in temperature.

### Step 5: Relate Moles to Mass
The number of moles \( n \) can be expressed in terms of mass \( m \) and molecular mass \( M \):
\[ n = \frac{m}{M} \]

### Step 6: Substitute \( C_v \)
For an ideal gas, the molar heat capacity at constant volume \( C_v \) can be expressed in terms of the specific heat ratio \( \gamma \):
\[ C_v = \frac{R}{\gamma - 1} \]

### Step 7: Equate Kinetic Energy and Internal Energy Change
Now we can equate the kinetic energy to the change in internal energy:
\[ \frac{1}{2} mv^2 = n C_v \Delta T \]
Substituting for \( n \) and \( C_v \):
\[ \frac{1}{2} mv^2 = \left(\frac{m}{M}\right) \left(\frac{R}{\gamma - 1}\right) \Delta T \]

### Step 8: Solve for \( \Delta T \)
Rearranging gives:
\[ \Delta T = \frac{mv^2 (\gamma - 1)}{2R} \]
Now, we can cancel \( m \) from both sides:
\[ \Delta T = \frac{v^2 (\gamma - 1)}{2R} \]

### Final Answer
Thus, the increase in temperature of the gas when the vessel is brought to rest is:
\[ \Delta T = \frac{v^2 (\gamma - 1)}{2R} \]