10 litre of a non linear polyatomic ideal gas at `127^(@)C` and 2 atm pressure is suddenly released to 1 atm pressure and the gas expanded adiabatically against constant external pressure. The final temperature and volume of the gas respectively are.

Options

A. T=350K,V = 17.5L
B. T = 300 K ,V = 15 L
C. T = 250 K, V = 12.5 L
D. None of these

Explanation

To solve the problem step-by-step, we will follow the principles of thermodynamics, particularly focusing on the adiabatic expansion of an ideal gas.

### Step 1: Identify Given Data
- Initial volume (V1) = 10 liters
- Initial temperature (T1) = 127°C = 127 + 273 = 400 K
- Initial pressure (P1) = 2 atm
- Final pressure (P2) = 1 atm
- The process is adiabatic.

### Step 2: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings (Q = 0). According to the first law of thermodynamics, the change in internal energy (ΔU) is equal to the work done (W) on or by the system.

### Step 3: Use the Formula for Adiabatic Process
For an ideal gas undergoing an adiabatic process, we can use the following relation:
\[ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}} \]
Where:
- \( \gamma \) (gamma) = \( \frac{C_p}{C_v} \) for the gas. For a non-linear polyatomic gas, \( \gamma \approx 1.33 \).

### Step 4: Calculate the Final Temperature (T2)
Substituting the known values into the equation:
\[ \frac{T_2}{400} = \left(\frac{1}{2}\right)^{\frac{1.33 - 1}{1.33}} \]

Calculating the exponent:
\[ \frac{1.33 - 1}{1.33} = \frac{0.33}{1.33} \approx 0.248 \]
Thus:
\[ \frac{T_2}{400} = \left(\frac{1}{2}\right)^{0.248} \]
Calculating \( \left(\frac{1}{2}\right)^{0.248} \):
\[ \left(\frac{1}{2}\right)^{0.248} \approx 0.84 \quad (\text{using a calculator}) \]
Now substituting back:
\[ T_2 = 400 \times 0.84 \approx 336 \text{ K} \]

### Step 5: Calculate the Final Volume (V2)
Using the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearranging gives:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]
Substituting the known values:
\[ V_2 = \frac{2 \, \text{atm} \times 10 \, \text{L} \times 336 \, \text{K}}{1 \, \text{atm} \times 400 \, \text{K}} \]
Calculating:
\[ V_2 = \frac{6720}{400} = 16.8 \, \text{L} \]

### Final Results
- Final Temperature (T2) = 336 K
- Final Volume (V2) = 16.8 L

### Summary
The final temperature and volume of the gas after adiabatic expansion are approximately \( T_2 = 336 \, \text{K} \) and \( V_2 = 16.8 \, \text{L} \).