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

Answer

To solve the problem, we need to analyze the situation where a box containing an ideal gas suddenly stops, and we want to find the change in temperature of the gas.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The box contains 2 moles of an ideal gas.
- The gas has a molar mass \( M \) and an adiabatic exponent \( \gamma \).
- The box is moving with a constant speed \( v \).

2. **Kinetic Energy of the Gas**:
- The kinetic energy \( KE \) of the gas when the box is moving can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
- Here, \( m \) is the mass of the gas. Since we have 2 moles of gas, the mass can be calculated as:
\[ m = n \cdot M = 2M \]
- Thus, the kinetic energy becomes:
\[ KE = \frac{1}{2} (2M) v^2 = Mv^2 \]

3. **Internal Energy of the Gas**:
- The internal energy \( U \) of an ideal gas is given by:
\[ U = nC_vT \]
- For 2 moles of gas, this becomes:
\[ U = 2C_vT \]

4. **Relating Internal Energy and Kinetic Energy**:
- When the box suddenly stops, the kinetic energy of the gas is converted into internal energy. Therefore, we can set the kinetic energy equal to the change in internal energy:
\[ Mv^2 = 2C_v \Delta T \]
- Rearranging gives us:
\[ \Delta T = \frac{Mv^2}{2C_v} \]

5. **Finding \( C_v \) in terms of \( R \) and \( \gamma \)**:
- We know that:
\[ C_p - C_v = R \]
- And from the definition of \( \gamma \):
\[ \gamma = \frac{C_p}{C_v} \]
- Rearranging gives:
\[ C_p = \gamma C_v \]
- Substituting into the first equation:
\[ \gamma C_v - C_v = R \implies C_v(\gamma - 1) = R \implies C_v = \frac{R}{\gamma - 1} \]

6. **Substituting \( C_v \) back into the equation for \( \Delta T \)**:
- We substitute \( C_v \) into the equation for \( \Delta T \):
\[ \Delta T = \frac{Mv^2}{2 \left(\frac{R}{\gamma - 1}\right)} = \frac{Mv^2 (\gamma - 1)}{2R} \]

7. **Final Expression for Change in Temperature**:
- Therefore, the change in temperature of the gas when the box suddenly stops is given by:
\[ \Delta T = \frac{Mv^2 (\gamma - 1)}{2R} \]