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.
(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

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: Understand the Initial Conditions
We have:
- Initial volume \( V_1 = 10 \, \text{L} \)
- Initial temperature \( T_1 = 127^\circ C = 127 + 273 = 400 \, \text{K} \)
- Initial pressure \( P_1 = 2 \, \text{atm} \)

### Step 2: Identify the Final Conditions
The gas expands to a final pressure of \( P_2 = 1 \, \text{atm} \). We need to find the final temperature \( T_2 \) and final volume \( V_2 \).

### Step 3: Use the First Law of Thermodynamics
Since the process is adiabatic, the heat exchange \( Q = 0 \). Therefore, the change in internal energy \( \Delta U \) is equal to the work done \( W \):
\[ \Delta U = W \]

### Step 4: Relate Change in Internal Energy to Temperature
For an ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = nC_v(T_2 - T_1) \]
Where \( C_v \) is the molar heat capacity at constant volume.

### Step 5: Use the Adiabatic Condition
For an adiabatic process, we can use the relation:
\[ \Delta U = -P_{\text{ext}}(V_2 - V_1) \]
Since the external pressure is constant at \( P_{\text{ext}} = 1 \, \text{atm} \).

### Step 6: Calculate the Molar Heat Capacity Ratio \( \gamma \)
For a non-linear polyatomic ideal gas, the ratio of specific heats \( \gamma \) is given as:
\[ \gamma = \frac{C_p}{C_v} = 1.33 \]

### Step 7: Express Work Done in Terms of Temperature and Pressure
Using the ideal gas law \( PV = nRT \), we can express the work done in terms of temperature:
\[ W = -P_{\text{ext}}(V_2 - V_1) = -P_{\text{ext}} \left( \frac{nRT_2}{P_2} - V_1 \right) \]

### Step 8: Substitute Values and Solve for \( T_2 \)
From the above equations, we can derive:
\[ nR\left(\frac{T_2}{P_2} - T_1\right) = -P_{\text{ext}}(V_2 - V_1) \]
Substituting \( P_2 = 1 \, \text{atm} \), \( V_1 = 10 \, \text{L} \), and \( T_1 = 400 \, \text{K} \), we can solve for \( T_2 \).

### Step 9: Solve for \( V_2 \)
Once we have \( T_2 \), we can find \( V_2 \) using the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearranging gives:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 10: Calculate Final Values
1. Calculate \( T_2 \) using the derived equations.
2. Calculate \( V_2 \) using the ideal gas law.

### Final Answer
After performing the calculations, we find:
- Final Temperature \( T_2 = 350 \, \text{K} \)
- Final Volume \( V_2 = 17.5 \, \text{L} \)

Thus, the correct answer is:
**(a) \( T = 350 \, \text{K}, V = 17.5 \, \text{L} \)**