Inversion temperature `(T_(i)=(2a)/(Rb))` is defined as the temperature above which if gas is expanded
adiabatically it gets warm up but if temperature of gas is lower than `T_(i)` then it will cool down. What will
happen to gas if it is adiabatically expanded at `50^(@)C` if its Boyle's temperature is `20^(@)C`

Options

A. Heating
B. Cooling
C. Constant
D. None

Answer

To solve the problem, we need to determine what happens to a gas when it is adiabatically expanded at a temperature of \(50^\circ C\) given that its Boyle's temperature is \(20^\circ C\).

### Step-by-Step Solution:

1. **Identify the Boyle's Temperature**:
- The Boyle's temperature (\(T_b\)) is given as \(20^\circ C\).
- Convert this temperature to Kelvin:
\[ T_b = 20 + 273 = 293 \, K \]

2. **Calculate the Inversion Temperature**:
- The inversion temperature (\(T_i\)) is defined as:
\[ T_i = \frac{2a}{Rb} \]
- It can also be expressed in terms of Boyle's temperature:
\[ T_i = 2 \times T_b \]
- Substitute the value of \(T_b\):
\[ T_i = 2 \times 293 \, K = 586 \, K \]
- Convert \(T_i\) back to Celsius:
\[ T_i = 586 - 273 = 313 \, ^\circ C \]

3. **Compare the Gas Temperature with Inversion Temperature**:
- The gas is at a temperature of \(50^\circ C\).
- Compare \(50^\circ C\) with \(313^\circ C\):
\[ 50^\circ C < 313^\circ C \]
- Since the gas temperature is lower than the inversion temperature, we can conclude that it will cool down when expanded adiabatically.

4. **Conclusion**:
- When the gas is adiabatically expanded at \(50^\circ C\), it will cool down.

### Final Answer:
The gas will cool down when adiabatically expanded at \(50^\circ C\).

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