1L of `NH_(3) " at " 27^(@)C` is expanded adiabatically to x litres and final temperature is `-123^(@)C`. What is the value of x?

Answer

To solve the problem of finding the final volume \( x \) of ammonia (\( NH_3 \)) after an adiabatic expansion, we will use the relation for an adiabatic process involving temperature and volume. Here’s a step-by-step solution:

### Step 1: Identify Initial Conditions
- **Initial Volume (\( V_1 \))**: 1 L
- **Initial Temperature (\( T_1 \))**: 27°C

### Step 2: Convert Temperatures to Kelvin
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 \):
\[ T_1 = 27 + 273 = 300 \, K \]
- For the final temperature \( T_2 = -123°C \):
\[ T_2 = -123 + 273 = 150 \, K \]

### Step 3: Use the Adiabatic Relation
In an adiabatic process, the following relation holds:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Where \( \gamma \) (gamma) is the heat capacity ratio. For ammonia (\( NH_3 \)), \( \gamma \) is approximately \( \frac{4}{3} \).

### Step 4: Calculate \( \gamma - 1 \)
\[ \gamma - 1 = \frac{4}{3} - 1 = \frac{1}{3} \]

### Step 5: Substitute Known Values into the Equation
Substituting the known values into the adiabatic relation:
\[ 300 \times 1^{\frac{1}{3}} = 150 \times V_2^{\frac{1}{3}} \]
This simplifies to:
\[ 300 = 150 \times V_2^{\frac{1}{3}} \]

### Step 6: Solve for \( V_2^{\frac{1}{3}} \)
Dividing both sides by 150 gives:
\[ 2 = V_2^{\frac{1}{3}} \]

### Step 7: Cube Both Sides to Find \( V_2 \)
Cubing both sides results in:
\[ V_2 = 2^3 = 8 \, L \]

### Conclusion
The final volume \( x \) after adiabatic expansion is:
\[ \boxed{8 \, L} \]