A gas consisting of a rigid diatomic molecules was initially under standard condition.Then,gas was compressed adiabatically to one fifth of its intitial volume What will be the mean kinetic energy of a rotating molecule in the final state?

Options

A. 1.44 J
B. 4.55 J
C. `787.98xx10^(-23) J`
D. `757.3xx10^(-23) J`

Answer

To solve the problem step by step, we need to follow these steps:

### Step 1: Understand the Initial Conditions
The gas is initially under standard conditions, which means:
- Initial temperature, \( T_1 = 273 \, \text{K} \)
- The gas is diatomic, which means it has two degrees of freedom for rotation.

### Step 2: Use the Adiabatic Process Relation
For an adiabatic process, the relation between temperature and volume is given by:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \( \gamma \) (gamma) for a diatomic gas is \( \frac{7}{5} \).

### Step 3: Determine the Final Volume
The gas is compressed to one-fifth of its initial volume:
\[ V_2 = \frac{1}{5} V_1 \]

### Step 4: Substitute the Values into the Adiabatic Relation
Substituting \( V_2 \) into the adiabatic relation:
\[ T_1 V_1^{\frac{7}{5} - 1} = T_2 \left(\frac{1}{5} V_1\right)^{\frac{7}{5} - 1} \]
This simplifies to:
\[ T_1 V_1^{\frac{2}{5}} = T_2 \left(\frac{1}{5^{\frac{2}{5}}} V_1^{\frac{2}{5}}\right) \]
Cancelling \( V_1^{\frac{2}{5}} \) from both sides gives:
\[ T_1 = T_2 \cdot 5^{\frac{2}{5}} \]

### Step 5: Solve for Final Temperature \( T_2 \)
Rearranging the equation gives:
\[ T_2 = \frac{T_1}{5^{\frac{2}{5}}} \]
Substituting \( T_1 = 273 \, \text{K} \):
\[ T_2 = \frac{273}{5^{\frac{2}{5}}} \]

### Step 6: Calculate \( 5^{\frac{2}{5}} \)
Calculating \( 5^{\frac{2}{5}} \):
\[ 5^{\frac{2}{5}} \approx 2.236 \]
Thus:
\[ T_2 \approx \frac{273}{2.236} \approx 122.4 \, \text{K} \]

### Step 7: Calculate the Mean Kinetic Energy
The mean kinetic energy of a rotating diatomic molecule is given by:
\[ KE = \frac{2}{2} k T \]
Where \( k \) is the Boltzmann constant \( k \approx 1.38 \times 10^{-23} \, \text{J/K} \).

Substituting \( T_2 \):
\[ KE = k T_2 = (1.38 \times 10^{-23}) \times (122.4) \]
Calculating this gives:
\[ KE \approx 1.69 \times 10^{-21} \, \text{J} \]

### Step 8: Final Result
The mean kinetic energy of a rotating molecule in the final state is approximately:
\[ KE \approx 1.69 \times 10^{-21} \, \text{J} \]