A gas (`gamma` = 1.5) is enclosed in a thermally insulated container of volume `4 xx 10^(-4) m^(-3)` 1 atmospheric pressure and at a temperature of 300K. If the gas is suddenly compressed to a volume of `10^(-4) m^3`, what will be the final pressure and the temperature of the gas. What will be your answers for final pressure and temperature if gas is slowly compressed to the same final volume.

Answer

To solve the problem, we will analyze the two scenarios: sudden compression and slow compression of the gas.

### Step 1: Understanding the Initial Conditions
We are given:
- \( \gamma = 1.5 \)
- Initial volume \( V_i = 4 \times 10^{-4} \, m^3 \)
- Final volume \( V_f = 10^{-4} \, m^3 \)
- Initial pressure \( P_i = 1 \, atm \) (which is approximately \( 101325 \, Pa \))
- Initial temperature \( T_i = 300 \, K \)

### Step 2: Sudden Compression (Adiabatic Process)
Since the gas is in a thermally insulated container, the process is adiabatic. For an adiabatic process, we can use the relation:
\[ P V^\gamma = \text{constant} \]
From this, we can write:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
Rearranging gives:
\[ P_f = P_i \left( \frac{V_i}{V_f} \right)^\gamma \]

### Step 3: Calculate Final Pressure \( P_f \)
Substituting the known values:
\[ P_f = 1 \, atm \left( \frac{4 \times 10^{-4}}{10^{-4}} \right)^{1.5} \]
Calculating the volume ratio:
\[ \frac{V_i}{V_f} = \frac{4 \times 10^{-4}}{10^{-4}} = 4 \]
Now substituting this back:
\[ P_f = 1 \, atm \times 4^{1.5} = 1 \, atm \times 8 = 8 \, atm \]

### Step 4: Calculate Final Temperature \( T_f \)
Using the relation for temperature in an adiabatic process:
\[ T_f = T_i \left( \frac{V_f}{V_i} \right)^{\gamma - 1} \]
Substituting the known values:
\[ T_f = 300 \, K \left( \frac{10^{-4}}{4 \times 10^{-4}} \right)^{0.5} \]
Calculating the volume ratio:
\[ \frac{V_f}{V_i} = \frac{10^{-4}}{4 \times 10^{-4}} = \frac{1}{4} \]
Now substituting this back:
\[ T_f = 300 \, K \times \left( \frac{1}{4} \right)^{0.5} = 300 \, K \times \frac{1}{2} = 150 \, K \]

### Step 5: Slow Compression
If the gas is slowly compressed, the process is still adiabatic because the container is thermally insulated. Thus, the calculations for final pressure and temperature remain the same:
- Final Pressure \( P_f = 8 \, atm \)
- Final Temperature \( T_f = 150 \, K \)

### Final Answers
- Final Pressure after sudden or slow compression: \( P_f = 8 \, atm \)
- Final Temperature after sudden or slow compression: \( T_f = 150 \, K \)