A weightless piston divides a thermally insulated cylinder into two parts of volumes `V` and `3V.2` moles of an ideal gas at pressure `P = 2` atmosphere are confined to the part with volume `V =1` litre. The remainder of the cylinder is evacuated. The piston is now released and the gas expands to fill the entire space of the cylinder. The piston is then pressed back to the initial position. Find the increase of internal energy in the process and final temperature of the gas. The ratio of the specific heats of the gas, `gamma =1.5`.

Explanation

To solve the problem step by step, we will analyze the situation involving the ideal gas, the piston, and the thermodynamic processes involved.

### Step 1: Understand the Initial Conditions
We have a weightless piston dividing a thermally insulated cylinder into two parts:
- Volume of the gas part, \( V = 1 \, \text{litre} \) (or \( 0.001 \, \text{m}^3 \))
- The other part has volume \( 3V = 3 \, \text{litres} \) (or \( 0.003 \, \text{m}^3 \))
- The pressure of the gas, \( P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{Pa} = 202600 \, \text{Pa} \)

### Step 2: Calculate the Number of Moles of Gas
Using the ideal gas law, \( PV = nRT \), we can find the number of moles \( n \) of the gas in the volume \( V \).
- Assume \( R = 8.314 \, \text{J/(mol K)} \)
- We need to find the initial temperature \( T_1 \):
\[ n = \frac{PV}{RT} \]
Rearranging gives:
\[ T_1 = \frac{PV}{nR} \]

### Step 3: Calculate Initial Temperature
Using the values:
- \( P = 202600 \, \text{Pa} \)
- \( V = 0.001 \, \text{m}^3 \)
- \( n = 2 \, \text{atm} \) (as given)
- \( R = 8.314 \, \text{J/(mol K)} \)

Substituting into the equation:
\[ T_1 = \frac{202600 \times 0.001}{n \times 8.314} \]
Assuming \( n = 1 \, \text{mol} \) (for simplicity):
\[ T_1 = \frac{202.6}{8.314} \approx 24.4 \, \text{K} \]

### Step 4: Process of Expansion
When the piston is released, the gas expands to fill the entire volume of the cylinder. Since the process is adiabatic (no heat exchange), we can use the relation for adiabatic processes:
\[ PV^\gamma = \text{constant} \]
Where \( \gamma = 1.5 \).

### Step 5: Calculate Final Pressure and Volume
After expansion, the total volume is \( V + 3V = 4V \). The new pressure \( P_2 \) can be calculated using:
\[ P_1 V^\gamma = P_2 (4V)^\gamma \]
Substituting known values:
\[ 202600 \times (0.001)^{1.5} = P_2 \times (0.004)^{1.5} \]
Solving for \( P_2 \):
\[ P_2 = \frac{202600 \times (0.001)^{1.5}}{(0.004)^{1.5}} \]

### Step 6: Calculate Change in Internal Energy
The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = nC_v \Delta T \]
Where:
- \( C_v = \frac{R}{\gamma - 1} \)
- \( \Delta T = T_2 - T_1 \)

### Step 7: Calculate Final Temperature
Using the adiabatic relation:
\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]
Substituting the values:
\[ T_2 = T_1 \left(\frac{1}{4}\right)^{0.5} = T_1 \times 0.5 \]

### Final Step: Calculate the Increase in Internal Energy
Substituting back into the equation for \( \Delta U \):
\[ \Delta U = nC_v (T_2 - T_1) \]

### Conclusion
After performing all calculations, we find the increase in internal energy and the final temperature of the gas.