A box filled with gas moving with constant velocity 30 m/s. Having monoatomic gas of mass (4u). Now block is suddenly stopped. Then find the change in temperature of gas

Options

A. 10K
B. 1K
C. 3K
D. 144K

Answer

To solve the problem, we need to find the change in temperature of a monoatomic gas when the box it is in is suddenly stopped from moving at a constant velocity of 30 m/s.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The box is moving with a velocity \( v = 30 \, \text{m/s} \).
- The mass of the gas is \( m = 4u \) (where \( u \) is the atomic mass unit).
- The gas is monoatomic.

2. **Calculate the Initial Kinetic Energy (KE)**:
The kinetic energy of the gas when it is moving can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times (4u) \times (30)^2 = 2u \times 900 = 1800u \, \text{J} \]

3. **Determine the Internal Energy of the Gas**:
For a monoatomic gas, the internal energy \( U \) can be expressed as:
\[ U = \frac{3}{2} nRT \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

4. **Use the Conservation of Energy**:
When the box is stopped, all the kinetic energy is converted into internal energy. Thus, we can set up the equation:
\[ KE + U_1 = U_2 \]
Since the box stops, the final kinetic energy \( KE_2 = 0 \). Therefore:
\[ 1800u + \frac{3}{2} nRT_1 = \frac{3}{2} nRT_2 \]

5. **Relate Moles to Mass**:
The number of moles \( n \) can be expressed in terms of mass \( m \) and molar mass \( M \):
\[ n = \frac{m}{M} \]
Given that the molar mass of the gas is \( 4u \), we have:
\[ n = \frac{4u}{4u} = 1 \, \text{mole} \]

6. **Substituting Values**:
Now substituting \( n = 1 \) mole into the energy equation:
\[ 1800u + \frac{3}{2} R T_1 = \frac{3}{2} R T_2 \]
Rearranging gives:
\[ 1800u = \frac{3}{2} R (T_2 - T_1) \]

7. **Calculate the Change in Temperature**:
Solving for \( \Delta T = T_2 - T_1 \):
\[ \Delta T = \frac{2 \times 1800u}{3R} \]
Using \( R = 8.314 \, \text{J/(mol K)} \):
\[ \Delta T = \frac{3600u}{3 \times 8.314} = \frac{3600u}{24.942} \approx 144.3 \, \text{K} \]

### Final Answer:
The change in temperature of the gas is approximately \( \Delta T \approx 144.3 \, \text{K} \).