An insulated box containing a monoatomic gas of molar mass (M) moving with a speed `v_0` is suddenly stopped. Find the increment is gas temperature as a result of stopping the box.

Answer

To solve the problem of finding the increment in gas temperature when an insulated box containing a monoatomic gas is suddenly stopped, we can follow these steps:

### Step 1: Understand the Problem
The insulated box contains a monoatomic gas moving with an initial speed \( v_0 \). When the box is stopped, the kinetic energy of the gas is converted into internal energy, which leads to an increase in temperature.

### Step 2: Calculate the Initial Kinetic Energy
The initial kinetic energy (KE) of the gas in the box can be expressed as:
\[ KE = \frac{1}{2} mv_0^2 \]
where \( m \) is the mass of the gas.

### Step 3: Relate Kinetic Energy to Internal Energy
Since the box is insulated, there is no heat exchange with the surroundings. The decrease in kinetic energy will result in an increase in internal energy (\( U \)) of the gas. The relationship can be expressed as:
\[ \Delta KE = \Delta U \]
Thus, we have:
\[ \frac{1}{2} mv_0^2 = \Delta U \]

### Step 4: Express Internal Energy in Terms of Temperature
For a monoatomic ideal gas, the internal energy can be expressed as:
\[ U = n \cdot \frac{f}{2} R T \]
where:
- \( n \) is the number of moles of the gas,
- \( f \) is the degrees of freedom (for a monoatomic gas, \( f = 3 \)),
- \( R \) is the universal gas constant,
- \( T \) is the temperature.

The change in internal energy (\( \Delta U \)) can be expressed as:
\[ \Delta U = n \cdot \frac{f}{2} R \Delta T \]

### Step 5: Substitute Values
Substituting \( f = 3 \) for a monoatomic gas, we get:
\[ \Delta U = n \cdot \frac{3}{2} R \Delta T \]

### Step 6: Relate Moles to Mass
The number of moles \( n \) can be expressed in terms of mass \( m \) and molar mass \( M \):
\[ n = \frac{m}{M} \]
Substituting this into the equation for \( \Delta U \):
\[ \Delta U = \frac{m}{M} \cdot \frac{3}{2} R \Delta T \]

### Step 7: Equate Kinetic Energy and Internal Energy Change
Now, we equate the expressions for \( \Delta U \):
\[ \frac{1}{2} mv_0^2 = \frac{m}{M} \cdot \frac{3}{2} R \Delta T \]

### Step 8: Solve for \( \Delta T \)
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0^2 = \frac{3}{2M} R \Delta T \]
Rearranging gives:
\[ \Delta T = \frac{M v_0^2}{3R} \]

### Final Result
The increment in gas temperature as a result of stopping the box is:
\[ \Delta T = \frac{M v_0^2}{3R} \]

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