`200 cm^(3)` of a gas is compressed to `100cm^(3)` at atmospheric pressure `(10^(6) "dyne"//cm^(2))`. Find the resultant pressure if the change is (i) slow (ii) sudden Take `gamma= 1.4`.

Explanation

To solve the problem step by step, we will analyze both scenarios: (i) slow compression and (ii) sudden compression.

### Given Data:
- Initial Volume, \( V_1 = 200 \, \text{cm}^3 \)
- Final Volume, \( V_2 = 100 \, \text{cm}^3 \)
- Initial Pressure, \( P_1 = 10^6 \, \text{dyne/cm}^2 \)
- \( \gamma = 1.4 \)

### Part (i): Slow Compression (Isothermal Process)

1. **Understanding the Process**:
- In a slow compression, the process is quasi-static, meaning the system is in equilibrium at all times. This results in an isothermal process (constant temperature).

2. **Using the Ideal Gas Law**:
- For an isothermal process, the relationship between pressure and volume is given by:
\[ P_1 V_1 = P_2 V_2 \]

3. **Rearranging the Equation**:
- We can rearrange this equation to find the final pressure \( P_2 \):
\[ P_2 = P_1 \frac{V_1}{V_2} \]

4. **Substituting the Values**:
- Plugging in the values:
\[ P_2 = 10^6 \, \text{dyne/cm}^2 \times \frac{200 \, \text{cm}^3}{100 \, \text{cm}^3} \]
\[ P_2 = 10^6 \, \text{dyne/cm}^2 \times 2 = 2 \times 10^6 \, \text{dyne/cm}^2 \]

5. **Converting to Atmospheres**:
- Since \( 1 \, \text{atm} = 10^6 \, \text{dyne/cm}^2 \):
\[ P_2 = 2 \, \text{atm} \]

### Part (ii): Sudden Compression (Adiabatic Process)

1. **Understanding the Process**:
- In a sudden compression, the gas does not have time to exchange heat with its surroundings, resulting in an adiabatic process.

2. **Using the Adiabatic Condition**:
- For an adiabatic process, the relationship between pressure and volume is given by:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

3. **Rearranging the Equation**:
- Rearranging to find \( P_2 \):
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]

4. **Substituting the Values**:
- Plugging in the values:
\[ P_2 = 10^6 \, \text{dyne/cm}^2 \left( \frac{200 \, \text{cm}^3}{100 \, \text{cm}^3} \right)^{1.4} \]
\[ P_2 = 10^6 \, \text{dyne/cm}^2 \times (2)^{1.4} \]

5. **Calculating \( (2)^{1.4} \)**:
- Using a calculator:
\[ (2)^{1.4} \approx 2.639 \]

6. **Final Calculation**:
- Therefore:
\[ P_2 = 10^6 \, \text{dyne/cm}^2 \times 2.639 \approx 2.639 \times 10^6 \, \text{dyne/cm}^2 \]

7. **Converting to Atmospheres**:
- Converting to atmospheres:
\[ P_2 \approx 2.639 \, \text{atm} \]

### Final Answers:
- (i) For slow compression, the resultant pressure is \( 2 \, \text{atm} \).
- (ii) For sudden compression, the resultant pressure is approximately \( 2.639 \, \text{atm} \).