A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. if container is suddenly stopped then change in temperature of the gas (R= gas contant ) is `(X)/(3R)` Value of x is ____

Answer

To solve the problem, we need to find the change in temperature of a monoatomic gas when the container it is in is suddenly stopped. The steps to solve this are as follows:

### Step 1: Identify the initial kinetic energy of the gas
The kinetic energy (KE) of the gas when the container is moving with velocity \( v = 30 \, \text{m/s} \) can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the gas. Given that the mass of the gas is \( 4 \, \text{u} \) (where \( 1 \, \text{u} \approx 1.66 \times 10^{-27} \, \text{kg} \)), we can substitute this into the equation.

### Step 2: Calculate the change in kinetic energy when the container is stopped
When the container is suddenly stopped, the kinetic energy of the gas becomes zero. Therefore, the change in kinetic energy (\( \Delta KE \)) is:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2} m v^2 - 0 = \frac{1}{2} m v^2 \]

### Step 3: Relate the change in kinetic energy to the change in temperature
For a monoatomic gas, the change in internal energy (\( \Delta U \)) is given by:
\[ \Delta U = n C_v \Delta T \]
where \( n \) is the number of moles, \( C_v \) is the molar heat capacity at constant volume, and \( \Delta T \) is the change in temperature. For a monoatomic gas, \( C_v = \frac{3R}{2} \).

### Step 4: Convert mass to moles
To find \( n \), the number of moles, we can use the formula:
\[ n = \frac{m}{M} \]
where \( M \) is the molar mass. Since the mass is given in atomic mass units (u), we can convert it to grams per mole. For a monoatomic gas, the molar mass \( M \) is approximately \( 4 \, \text{g/mol} \) (since \( 4 \, \text{u} \) corresponds to \( 4 \, \text{g/mol} \)).

### Step 5: Substitute into the energy equation
Now we can substitute \( n \) and \( C_v \) into the energy equation:
\[ \Delta KE = n C_v \Delta T \]
Substituting \( n = \frac{4}{4} = 1 \, \text{mol} \) and \( C_v = \frac{3R}{2} \):
\[ \Delta KE = 1 \cdot \frac{3R}{2} \Delta T \]

### Step 6: Solve for \( \Delta T \)
Now we can solve for \( \Delta T \):
\[ \Delta T = \frac{\Delta KE}{\frac{3R}{2}} = \frac{2 \Delta KE}{3R} \]

### Step 7: Substitute \( \Delta KE \) into the equation
Substituting \( \Delta KE = \frac{1}{2} m v^2 \):
\[ \Delta T = \frac{2 \cdot \frac{1}{2} m v^2}{\frac{3R}{2}} = \frac{m v^2}{3R} \]

### Step 8: Substitute the values
Now substituting \( m = 4 \, \text{g/mol} \) and \( v = 30 \, \text{m/s} \):
\[ \Delta T = \frac{4 \cdot (30)^2}{3R} = \frac{4 \cdot 900}{3R} = \frac{3600}{3R} \]

### Step 9: Finalize the answer
Thus, the change in temperature of the gas is:
\[ \Delta T = \frac{3600}{3R} \]
From the problem statement, we have \( \Delta T = \frac{X}{3R} \), which gives us \( X = 3600 \).

### Final Answer
The value of \( X \) is **3600**.