When `1mol` of a monoatomic ideal gas at `TK` undergoes adiabatic change under a constant external pressure of `1atm`, changes volume from `1 L to 2L`. The final temperature (in K) would be

Options

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

Explanation

To find the final temperature of the monoatomic ideal gas after an adiabatic change, we can follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings (Q = 0). Therefore, the change in internal energy (ΔU) is equal to the work done on the system (W).

### Step 2: Write the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q + W \]
Since \( Q = 0 \) for an adiabatic process:
\[ \Delta U = W \]

### Step 3: Express Change in Internal Energy
For an ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_v \Delta T \]
Where:
- \( n \) = number of moles (1 mol)
- \( C_v \) = molar heat capacity at constant volume
- \( \Delta T = T_f - T_i \) (final temperature - initial temperature)

### Step 4: Express Work Done
The work done on the gas during an adiabatic process can be expressed as:
\[ W = -P \Delta V \]
Where:
- \( P \) = external pressure (1 atm)
- \( \Delta V = V_f - V_i = 2L - 1L = 1L \)

### Step 5: Calculate Work Done
Convert 1 L to cubic meters for SI units:
\[ 1L = 0.001 m^3 \]
Thus:
\[ \Delta V = 1L = 0.001 m^3 \]
Now, substituting the values:
\[ W = -P \Delta V = -1 \text{ atm} \times 0.001 m^3 \]
Convert 1 atm to Joules:
\[ 1 \text{ atm} = 101.325 \text{ J} \]
So:
\[ W = -101.325 \times 0.001 = -0.101325 \text{ J} \]

### Step 6: Relate ΔU and W
From the first law:
\[ n C_v (T_f - T_i) = W \]
For a monoatomic ideal gas, \( C_v = \frac{3}{2} R \), where \( R = 8.314 \text{ J/(mol K)} \).
Thus:
\[ C_v = \frac{3}{2} \times 8.314 = 12.471 \text{ J/(mol K)} \]

### Step 7: Substitute Values
Now substituting the values into the equation:
\[ 1 \times 12.471 (T_f - T) = -0.101325 \]
This simplifies to:
\[ 12.471 (T_f - T) = -0.101325 \]

### Step 8: Solve for Final Temperature
Rearranging gives:
\[ T_f - T = \frac{-0.101325}{12.471} \]
Calculating the right side:
\[ T_f - T \approx -0.00812 \]
Thus:
\[ T_f \approx T - 0.00812 \]

### Final Result
The final temperature \( T_f \) in Kelvin can be expressed as:
\[ T_f = T - 0.00812 \text{ K} \]