A quantity of air at normal temperature is compressed (a) slowly (b) suddenly to one third of its volume. Find the rise in temperature, if any in each case, `gamma= 1.4`.

Explanation

To solve the problem, we need to analyze two scenarios of compressing air: (a) slowly and (b) suddenly. We will calculate the rise in temperature for each case using the principles of thermodynamics.

### Step-by-Step Solution:

#### (a) Slow Compression (Isothermal Process)

1. **Understanding the Process**: When air is compressed slowly, it is a quasi-static process. In this case, the system is in thermal equilibrium with its surroundings, allowing heat exchange.

2. **Temperature Change**: Since the process is isothermal (constant temperature), there is no change in temperature during slow compression.

\[ \Delta T = 0 \]

3. **Conclusion for Slow Compression**: The rise in temperature for the slow compression of air is zero.

#### (b) Sudden Compression (Adiabatic Process)

1. **Understanding the Process**: When air is compressed suddenly, there is no time for heat exchange with the surroundings. This makes the process adiabatic, meaning that no heat is transferred into or out of the system.

2. **Using the Adiabatic Relation**: For an adiabatic process, the relationship between temperature and volume is given by:

\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

where \( \gamma = 1.4 \).

3. **Initial and Final Volumes**: We are compressing the air to one third of its volume. Thus, if \( V_2 = \frac{V_1}{3} \), we can substitute this into the equation.

4. **Rearranging the Equation**: We want to find \( T_2 \):

\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} \]

Substituting \( V_2 = \frac{V_1}{3} \):

\[ T_2 = T_1 \left( \frac{V_1}{\frac{V_1}{3}} \right)^{\gamma - 1} = T_1 \left( 3 \right)^{\gamma - 1} \]

5. **Calculating \( T_2 \)**: Since \( \gamma - 1 = 0.4 \):

\[ T_2 = T_1 \cdot 3^{0.4} \]

Using \( T_1 = 273 \, \text{K} \):

\[ T_2 = 273 \cdot 3^{0.4} \]

Calculating \( 3^{0.4} \approx 1.5157 \):

\[ T_2 \approx 273 \cdot 1.5157 \approx 414.5 \, \text{K} \]

6. **Finding the Rise in Temperature**: The rise in temperature \( \Delta T \) is given by:

\[ \Delta T = T_2 - T_1 = 414.5 - 273 \approx 141.5 \, \text{K} \]

### Final Results:
- For slow compression: \( \Delta T = 0 \, \text{K} \)
- For sudden compression: \( \Delta T \approx 141.5 \, \text{K} \)