When one mole of a monoatomic ideal gas at initial temperature T K expands adiabatically from 1 litre to 2litres , the final temperature in Kelvin would be

Options

A. T
B. `(T)/( 2^(2//3))`
C. `T - (2)/( 3xx 0.0821)`
D. `T + (2)/( 3 xx 0.0821)`

Explanation

To solve the problem of finding the final temperature of a monoatomic ideal gas after an adiabatic expansion, we can follow these steps:

### Step 1: Understand the Given Data
We have:
- Initial temperature \( T_1 = T \) (in Kelvin)
- Initial volume \( V_1 = 1 \, \text{L} \)
- Final volume \( V_2 = 2 \, \text{L} \)
- For a monoatomic ideal gas, the specific heat capacities are:
- \( C_V = \frac{3R}{2} \)
- \( C_P = \frac{5R}{2} \)

### Step 2: Calculate the Heat Capacity Ratio \( \gamma \)
The heat capacity ratio \( \gamma \) is given by:
\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{5R}{2}}{\frac{3R}{2}} = \frac{5}{3} \]

### Step 3: Use the Adiabatic Condition
For an adiabatic process, the relationship between the temperatures and volumes is given by:
\[ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ \frac{T_2}{T} = \left( \frac{1}{2} \right)^{\frac{5}{3} - 1} \]
This simplifies to:
\[ \frac{T_2}{T} = \left( \frac{1}{2} \right)^{\frac{2}{3}} \]

### Step 5: Solve for \( T_2 \)
Now we can express \( T_2 \):
\[ T_2 = T \left( \frac{1}{2} \right)^{\frac{2}{3}} \]

### Step 6: Final Expression
Thus, the final temperature \( T_2 \) can be expressed as:
\[ T_2 = T \cdot \frac{1}{\sqrt[3]{4}} = \frac{T}{\sqrt[3]{4}} \]

### Conclusion
The final temperature after the adiabatic expansion is:
\[ T_2 = T \cdot \left( \frac{1}{2} \right)^{\frac{2}{3}} \]