If at `50^(@)C` and `75 cm` of mercury pressure, a definite mass of gas is compressed (i) slowly (ii) suddenly, than what will be the final pressure and temp. of the gas in each case, if the final volume is one fourth of the initial volume? `gamma= 1.5`.

Explanation

To solve the problem step by step, we will analyze both cases: the slow compression (isothermal process) and the sudden compression (adiabatic process).

### Given Data:
- Initial Temperature, \( T_1 = 50^\circ C = 50 + 273 = 323 \, K \)
- Initial Pressure, \( P_1 = 75 \, cm \, of \, Hg \)
- Final Volume, \( V_2 = \frac{V_1}{4} \)
- \( \gamma = 1.5 \)

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

1. **Identify the relationship for isothermal process:**
In an isothermal process, the product of pressure and volume remains constant:
\[ P_1 V_1 = P_2 V_2 \]

2. **Substituting the final volume:**
Since \( V_2 = \frac{V_1}{4} \), we can rewrite the equation:
\[ P_1 V_1 = P_2 \left(\frac{V_1}{4}\right) \]

3. **Solving for final pressure \( P_2 \):**
Rearranging gives:
\[ P_2 = P_1 \times \frac{V_1}{V_2} = P_1 \times 4 \]
Substituting \( P_1 = 75 \, cm \, of \, Hg \):
\[ P_2 = 75 \times 4 = 300 \, cm \, of \, Hg \]

4. **Final Temperature \( T_2 \):**
In an isothermal process, the temperature remains constant:
\[ T_2 = T_1 = 50^\circ C \]

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

1. **Identify the relationship for adiabatic process:**
For an adiabatic process, we use the formula:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]

2. **Substituting the final volume:**
Again, since \( V_2 = \frac{V_1}{4} \):
\[ P_1 V_1^\gamma = P_2 \left(\frac{V_1}{4}\right)^\gamma \]

3. **Solving for final pressure \( P_2 \):**
Rearranging gives:
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P_1 \times 4^\gamma \]
Substituting \( P_1 = 75 \, cm \, of \, Hg \) and \( \gamma = 1.5 \):
\[ P_2 = 75 \times 4^{1.5} = 75 \times 8 = 600 \, cm \, of \, Hg \]

4. **Final Temperature \( T_2 \):**
For the adiabatic process, we use:
\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]
Substituting the values:
\[ T_2 = 323 \times 4^{1.5 - 1} = 323 \times 4^{0.5} = 323 \times 2 = 646 \, K \]
Converting back to Celsius:
\[ T_2 = 646 - 273 = 373^\circ C \]

### Final Results:
- **Slow Compression (Isothermal):**
- Final Pressure \( P_2 = 300 \, cm \, of \, Hg \)
- Final Temperature \( T_2 = 50^\circ C \)

- **Sudden Compression (Adiabatic):**
- Final Pressure \( P_2 = 600 \, cm \, of \, Hg \)
- Final Temperature \( T_2 = 373^\circ C \)