One mole of an ideal monoatomic gas at temperature `T` and volume `1L` expands to `2L` against a constant external pressure of one atm under adiabatic conditions, then final temperature of gas will be:

Options

A. `T+(2)/(3xx0.0821)`
B. `T-(2)/(3xx0.0821)`
C. `(T)/(2^(5//3-1))`
D. `(T)/(2^(5//3+1))`

Explanation

To solve the problem step by step, we will use the principles of thermodynamics, particularly focusing on the first law of thermodynamics and the properties of an ideal gas under adiabatic conditions.

### Step 1: Understand the Conditions
We have one mole of an ideal monoatomic gas that expands from a volume of 1 L to 2 L against a constant external pressure of 1 atm. The process is adiabatic, meaning no heat is exchanged with the surroundings (Q = 0).

**Hint:** Remember that in an adiabatic process, the change in internal energy (ΔU) is equal to the work done on or by the system.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Since the process is adiabatic, \( Q = 0 \), so:
\[ \Delta U = -W \]

### Step 3: Calculate Work Done (W)
For an expansion against a constant external pressure, the work done (W) is given by:
\[ W = -P_{\text{external}} \times (V_2 - V_1) \]
Where:
- \( P_{\text{external}} = 1 \, \text{atm} \)
- \( V_2 = 2 \, \text{L} \)
- \( V_1 = 1 \, \text{L} \)

Substituting the values:
\[ W = -1 \, \text{atm} \times (2 \, \text{L} - 1 \, \text{L}) = -1 \, \text{atm} \times 1 \, \text{L} = -1 \, \text{L atm} \]

**Hint:** Remember to convert the work done into Joules if needed. \( 1 \, \text{L atm} = 101.325 \, \text{J} \).

### Step 4: Relate Work Done to Change in Internal Energy
For an ideal monoatomic gas, the change in internal energy (ΔU) can be expressed as:
\[ \Delta U = n C_V \Delta T \]
Where:
- \( n = 1 \, \text{mol} \)
- \( C_V = \frac{3}{2} R \) (for a monoatomic gas)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)

Substituting \( C_V \):
\[ C_V = \frac{3}{2} \times 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} = 0.12315 \, \text{L atm K}^{-1} \text{mol}^{-1} \]

### Step 5: Set Up the Equation
From the first law:
\[ -W = n C_V \Delta T \]
Substituting the values:
\[ 1 \, \text{L atm} = 1 \times 0.12315 \, \Delta T \]

### Step 6: Solve for ΔT
Now we can solve for \( \Delta T \):
\[ \Delta T = \frac{1 \, \text{L atm}}{0.12315} \]

### Step 7: Calculate Final Temperature
The change in temperature \( \Delta T \) is related to the initial temperature \( T \):
\[ T_2 = T - \Delta T \]
Substituting \( \Delta T \):
\[ T_2 = T - \frac{1}{0.12315} \]

### Final Result
Now, calculate the numerical value of \( T_2 \):
\[ T_2 = T - 8.12 \, \text{K} \]

### Summary
The final temperature \( T_2 \) of the gas after expansion is:
\[ T_2 = T - 8.12 \, \text{K} \]