A box of negligible mass containing 2 moles of an ideal gas of molar mass M and adiabatic exponent `gamma` moves with constant speed v on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be :

Options

A. `((gamma-1) Mv^2)/(4R)`
B. `(gamma Mv^2)/(2R)`
C. `(Mv^2)/(2 (gamma -1 ) R)`
D. `((gamma -1) Mv^2)/(2R)`

Answer

To solve the problem, we will analyze the situation step by step:

### Step 1: Understand the scenario
We have a box with negligible mass containing 2 moles of an ideal gas. The box is moving with a constant speed \( v \) on a smooth horizontal surface. When the box suddenly stops, we need to determine the change in temperature of the gas inside the box.

### Step 2: Identify the energy transformation
When the box stops, the kinetic energy of the box is converted into the internal energy of the gas. This is because the gas will expand or contract due to the change in motion of the box, which affects its temperature.

### Step 3: Calculate the initial kinetic energy
The kinetic energy (\( KE \)) of the box can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Since the box has negligible mass, we can consider the effective mass of the gas, which is related to the number of moles. The total kinetic energy of the gas when the box is moving is:
\[ KE = 2 \times \frac{1}{2} mv^2 = mv^2 \]
where \( m \) is the effective mass of the gas.

### Step 4: Relate kinetic energy to change in internal energy
The change in internal energy (\( \Delta U \)) of the gas can be expressed in terms of the change in temperature (\( \Delta T \)):
\[ \Delta U = nC_v \Delta T \]
For an ideal gas, we can also express the internal energy in terms of the gas constant \( R \) and the adiabatic exponent \( \gamma \):
\[ \Delta U = nR \Delta T / (\gamma - 1) \]
where \( n \) is the number of moles of the gas.

### Step 5: Set kinetic energy equal to change in internal energy
Since the kinetic energy is converted into internal energy, we can set the two equations equal to each other:
\[ mv^2 = \frac{nR \Delta T}{\gamma - 1} \]
Substituting \( n = 2 \) (since we have 2 moles of gas):
\[ mv^2 = \frac{2R \Delta T}{\gamma - 1} \]

### Step 6: Solve for change in temperature
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{mv^2 (\gamma - 1)}{2R} \]

### Final Answer
Thus, the change in temperature of the gas when the box stops is:
\[ \Delta T = \frac{mv^2 (\gamma - 1)}{2R} \]