A quantity of air at `27^(@)C` and atmospheric pressure is suddenly compressed to half its original volume. Find the final (i) pressure and (ii) temperature.
Given `gamma for air = 1.42`.

Explanation

To solve the problem step by step, we will analyze the adiabatic process of compressing air and use the relevant equations.

### Given Data:
- Initial temperature, \( T_1 = 27^\circ C = 273 + 27 = 300 \, K \)
- Initial pressure, \( P_1 = 1 \, atm \)
- Initial volume, \( V_1 \)
- Final volume, \( V_2 = \frac{V_1}{2} \)
- \( \gamma \) (gamma) for air = 1.42

### Step 1: Find the Final Pressure \( P_2 \)

In an adiabatic process, the relationship between pressure and volume is given by the equation:

\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

Rearranging this equation to find \( P_2 \):

\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

Since \( V_2 = \frac{V_1}{2} \), we have:

\[ \frac{V_1}{V_2} = \frac{V_1}{\frac{V_1}{2}} = 2 \]

Substituting the values into the equation for \( P_2 \):

\[ P_2 = P_1 \cdot (2)^\gamma = 1 \, atm \cdot (2)^{1.42} \]

Calculating \( (2)^{1.42} \):

\[ (2)^{1.42} \approx 2.675 \]

Thus,

\[ P_2 \approx 1 \, atm \cdot 2.675 \approx 2.675 \, atm \]

### Step 2: Find the Final Temperature \( T_2 \)

In an adiabatic process, the relationship between temperature and volume is given by the equation:

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

Rearranging this equation to find \( T_2 \):

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

Substituting \( V_2 = \frac{V_1}{2} \):

\[ T_2 = T_1 \cdot \left( \frac{V_1}{\frac{V_1}{2}} \right)^{\gamma - 1} = T_1 \cdot (2)^{\gamma - 1} \]

Calculating \( \gamma - 1 \):

\[ \gamma - 1 = 1.42 - 1 = 0.42 \]

Now substituting \( T_1 = 300 \, K \):

\[ T_2 = 300 \cdot (2)^{0.42} \]

Calculating \( (2)^{0.42} \):

\[ (2)^{0.42} \approx 1.348 \]

Thus,

\[ T_2 \approx 300 \cdot 1.348 \approx 404.4 \, K \]

Converting \( T_2 \) back to Celsius:

\[ T_2 = 404.4 - 273 \approx 131.4^\circ C \]

### Final Results:
- Final Pressure \( P_2 \approx 2.675 \, atm \)
- Final Temperature \( T_2 \approx 131.4^\circ C \)