An insulated container containing monoatomic gas of molar mass m is moving with a velocity `V_(0)`. If the container is suddenly stopped , find the change in temperature .

Options

A. `(mv_(0)^(2))/(2R)`
B. `(mv(0)^(2))/(3R)`
C. `(R )/(mv_(0)^(2))`
D. `(3mv_(0)^(2))/(2R)`

Explanation

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

### Step 1: Understand the Initial Conditions
The container is moving with a velocity \( V_0 \) and contains a monoatomic gas with molar mass \( m \). When the container is stopped, the gas molecules will lose their kinetic energy associated with the motion of the container.

### Step 2: Calculate the Initial Kinetic Energy
The kinetic energy of the gas molecules due to the velocity \( V_0 \) can be expressed as:
\[ KE = \frac{1}{2} m n V_0^2 \]
where \( n \) is the number of moles of the gas.

### Step 3: Relate Kinetic Energy to Temperature Change
For a monoatomic ideal gas, the change in translational kinetic energy can also be related to the change in temperature using the equation:
\[ \Delta KE = \frac{3}{2} n R \Delta T \]
where \( R \) is the universal gas constant and \( \Delta T \) is the change in temperature.

### Step 4: Set the Kinetic Energy Equations Equal
Since the kinetic energy lost by the gas is equal to the increase in internal energy (which is related to the temperature change), we can set the two expressions for kinetic energy equal to each other:
\[ \frac{1}{2} m n V_0^2 = \frac{3}{2} n R \Delta T \]

### Step 5: Simplify the Equation
We can cancel \( n \) from both sides (assuming \( n \neq 0 \)):
\[ \frac{1}{2} m V_0^2 = \frac{3}{2} R \Delta T \]

### Step 6: Solve for Change in Temperature
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{m V_0^2}{3R} \]

### Final Answer
Thus, the change in temperature when the container is suddenly stopped is:
\[ \Delta T = \frac{m V_0^2}{3R} \]

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