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 after moving with a velocity of 200 m/s. Here’s a step-by-step solution:

### Step 1: Understand 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 Molecular Mass to Kilograms
Since we need to work in SI units, we convert the molecular mass from grams to kilograms:
\[ M = 40 \, \text{g/mol} = 0.040 \, \text{kg/mol} \]

### Step 3: Calculate the Kinetic Energy of the Gas
The kinetic energy (KE) of the gas can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
To find the mass \( m \) of the gas, we need to know the number of moles \( n \):
\[ m = n \cdot M = n \cdot 0.040 \, \text{kg} \]
Thus, the kinetic energy becomes:
\[ KE = \frac{1}{2} (n \cdot 0.040) (200)^2 \]
Calculating this gives:
\[ KE = \frac{1}{2} (n \cdot 0.040) (40000) = 20 n \, \text{J} \]

### Step 4: Relate Kinetic Energy to Internal Energy Change
When the gas is stopped, the kinetic energy is converted into internal energy. The change in internal energy (\( \Delta U \)) for a diatomic gas can be expressed as:
\[ \Delta U = n C_V \Delta T \]
For a diatomic gas, the molar heat capacity at constant volume \( C_V \) is given by:
\[ C_V = \frac{5}{2} R \]
Thus,
\[ \Delta U = n \left(\frac{5}{2} R\right) \Delta T \]

### Step 5: Set Kinetic Energy Equal to Change in Internal Energy
Since the kinetic energy is converted to internal energy, we have:
\[ 20n = n \left(\frac{5}{2} R\right) \Delta T \]
Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ 20 = \frac{5}{2} R \Delta T \]

### Step 6: Solve for \( \Delta T \)
Rearranging the equation gives:
\[ \Delta T = \frac{20 \cdot 2}{5R} = \frac{40}{5R} = \frac{8}{R} \]

### Step 7: Final Expression for Temperature Rise
Thus, the rise in temperature of the gas when it is suddenly stopped is:
\[ \Delta T = \frac{8}{R} \, \text{K} \]

### Step 8: Convert to Celsius
Since the change in temperature in Kelvin is equal to the change in temperature in Celsius, we have:
\[ \Delta T = \frac{8}{R} \, \text{°C} \]

### Conclusion
The rise in temperature of the gas is \( \Delta T = \frac{8}{R} \, \text{°C} \).

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