A gas (`gamma`=1.5) is enclosed in a container of volume `150 cm^3`. The initial pressure and the initial temperature are `1.5 xx 10^5` Pa and 300K respectively. Of The gas is adiabatically compressed to `50 cm^3`, find the final pressure and temperature and the work done by the gas in the process. Also find the total change in internal energy of the gas in the process.

Answer

To solve the problem step by step, we will follow the principles of thermodynamics, particularly focusing on the adiabatic process for an ideal gas.

### Given Data:
- \( \gamma = 1.5 \)
- Initial Volume \( V_i = 150 \, \text{cm}^3 = 150 \times 10^{-6} \, \text{m}^3 \)
- Final Volume \( V_f = 50 \, \text{cm}^3 = 50 \times 10^{-6} \, \text{m}^3 \)
- Initial Pressure \( P_i = 1.5 \times 10^5 \, \text{Pa} \)
- Initial Temperature \( T_i = 300 \, \text{K} \)

### Step 1: Calculate the Final Pressure \( P_f \)

Using the adiabatic condition for an ideal gas:
\[ P_i V_i^\gamma = P_f V_f^\gamma \]
Rearranging for \( P_f \):
\[ P_f = P_i \left( \frac{V_i}{V_f} \right)^\gamma \]

Substituting the values:
\[ P_f = 1.5 \times 10^5 \left( \frac{150 \times 10^{-6}}{50 \times 10^{-6}} \right)^{1.5} \]
\[ = 1.5 \times 10^5 \left( 3 \right)^{1.5} \]
Calculating \( 3^{1.5} \):
\[ 3^{1.5} = 3 \sqrt{3} \approx 5.196 \]
Thus,
\[ P_f = 1.5 \times 10^5 \times 5.196 \approx 7.794 \times 10^5 \, \text{Pa} \approx 7.8 \times 10^5 \, \text{Pa} \]

### Step 2: Calculate the Final Temperature \( T_f \)

Using the relation for adiabatic processes:
\[ \frac{T_f}{T_i} = \left( \frac{V_i}{V_f} \right)^{\gamma - 1} \]
Rearranging for \( T_f \):
\[ T_f = T_i \left( \frac{V_i}{V_f} \right)^{\gamma - 1} \]

Substituting the values:
\[ T_f = 300 \left( \frac{150 \times 10^{-6}}{50 \times 10^{-6}} \right)^{1.5 - 1} \]
\[ = 300 \left( 3 \right)^{0.5} \]
Calculating \( 3^{0.5} \):
\[ 3^{0.5} = \sqrt{3} \approx 1.732 \]
Thus,
\[ T_f = 300 \times 1.732 \approx 519.6 \, \text{K} \approx 520 \, \text{K} \]

### Step 3: Calculate the Work Done \( W \)

For an adiabatic process, the work done can be calculated using:
\[ W = \Delta U = nC_v(T_f - T_i) \]
Where \( C_v = \frac{R}{\gamma - 1} \).

However, we can also use the relation:
\[ W = \frac{P_i V_i - P_f V_f}{\gamma - 1} \]

Substituting the values:
\[ W = \frac{1.5 \times 10^5 \times 150 \times 10^{-6} - 7.8 \times 10^5 \times 50 \times 10^{-6}}{1.5 - 1} \]
Calculating each term:
\[ = \frac{(1.5 \times 150) - (7.8 \times 50)}{0.5} \times 10^{-1} \]
\[ = \frac{225 - 390}{0.5} \times 10^{-1} \]
\[ = \frac{-165}{0.5} \times 10^{-1} = -330 \, \text{J} \]

### Step 4: Calculate the Change in Internal Energy \( \Delta U \)

Since the process is adiabatic:
\[ \Delta U = W \]
Thus,
\[ \Delta U = -330 \, \text{J} \]

### Final Results:
- Final Pressure \( P_f \approx 7.8 \times 10^5 \, \text{Pa} \)
- Final Temperature \( T_f \approx 520 \, \text{K} \)
- Work Done \( W \approx -330 \, \text{J} \)
- Change in Internal Energy \( \Delta U \approx -330 \, \text{J} \)