A diatomic gas of molecular mass 40 g/mol is filled in rigid container at temperature `30^@ C`. It is moving with velocity 200 m/s. If its is suddenly stopped, the rise in the temperature of the gas is

Options

A. `(32)/(R ) ""^(@) C`
B. `(320)/(R ) ""^(@) C`
C. `(3200)/( R )""^(@ ) C`
D. `(3.2)/(R ) ""^@ C`

Answer

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

### Step 1: Identify the given data
- Molecular mass of the gas, \( M = 40 \, \text{g/mol} \)
- Initial temperature, \( T_i = 30^\circ C = 303 \, \text{K} \)
- Velocity of the gas, \( v = 200 \, \text{m/s} \)

### Step 2: Convert the molecular mass to kg
Since we need to work in SI units, we convert the molecular mass from grams to kilograms:
\[ M = 40 \, \text{g/mol} = 40 \times 10^{-3} \, \text{kg/mol} \]

### Step 3: Calculate the kinetic energy of the gas
The kinetic energy \( KE \) of the gas can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
Where \( m \) is the mass of the gas in kg. If we have \( N \) moles of gas, then the total mass \( m \) is:
\[ m = N \times M = N \times 40 \times 10^{-3} \, \text{kg} \]
Thus, the kinetic energy becomes:
\[ KE = \frac{1}{2} (N \times 40 \times 10^{-3}) v^2 \]
Substituting \( v = 200 \, \text{m/s} \):
\[ KE = \frac{1}{2} (N \times 40 \times 10^{-3}) (200)^2 \]
\[ KE = \frac{1}{2} (N \times 40 \times 10^{-3}) \times 40000 \]
\[ KE = 800N \, \text{J} \]

### Step 4: Relate kinetic energy to internal energy change
When the gas is suddenly stopped, the kinetic energy is converted to internal energy. For a diatomic gas, the change in internal energy \( \Delta U \) is given by:
\[ \Delta U = \frac{F}{2} N R \Delta T \]
Where:
- \( F = 5 \) (degrees of freedom for a diatomic gas)
- \( R = 8.314 \, \text{J/(mol K)} \)

### Step 5: Set kinetic energy equal to change in internal energy
Setting the kinetic energy equal to the change in internal energy:
\[ 800N = \frac{5}{2} N R \Delta T \]
We can cancel \( N \) from both sides (assuming \( N \neq 0 \)):
\[ 800 = \frac{5}{2} R \Delta T \]

### Step 6: Solve for \( \Delta T \)
Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{800 \times 2}{5R} \]
Substituting \( R = 8.314 \):
\[ \Delta T = \frac{1600}{5 \times 8.314} \]
\[ \Delta T = \frac{1600}{41.57} \approx 38.5 \, \text{K} \]

### Step 7: Conclusion
The rise in temperature of the gas is approximately \( 38.5 \, \text{K} \).