A cylinderical vessel made of thermally isulating material is divided into two equal parts by an insulating (moveable & smooth) piston. Both parts hae same ideal gas `((V_(p))/(C_(v))=(3)/(2))`. The gas in left part is supplied heat such that volume of right part becomes one fourth of initial value.
`(PVT,PVT)`
(1) In right part, the pressure becomes eight times of its inital value
(2) In right part, the temperature becomes two times
(3) In right part, work done by gas =3PV
(4) In right part, work done by gas=-2PV

Answer

To solve the problem step by step, we will analyze the situation involving the ideal gas in the cylindrical vessel divided by a piston.

### Step 1: Understand the initial conditions
- We have a cylindrical vessel divided into two equal parts by a movable piston.
- The gas in both parts is an ideal gas with the ratio \( \frac{V_p}{C_v} = \frac{3}{2} \).
- Let the initial pressure and volume in both parts be \( P_0 \) and \( V_0 \) respectively.

**Hint:** Identify the initial state of the gas and the properties of the ideal gas.

### Step 2: Analyze the changes in the right part
- The volume of the right part becomes one-fourth of its initial value, so the new volume \( V_2 = \frac{V_0}{4} \).
- Since the vessel is thermally insulating, the process in the right part is adiabatic.

**Hint:** Recognize that an adiabatic process means no heat exchange occurs.

### Step 3: Apply the adiabatic condition
- For an adiabatic process, we can use the relation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
where \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \) (since \( \frac{V_p}{C_v} = \frac{3}{2} \) implies \( C_v = \frac{3}{2}R \) and \( C_p = \frac{5}{2}R \)).

- Plugging in the values:
\[ P_0 V_0^{\frac{5}{3}} = P_2 \left(\frac{V_0}{4}\right)^{\frac{5}{3}} \]

**Hint:** Use the adiabatic condition to relate the pressures and volumes before and after the change.

### Step 4: Solve for the new pressure \( P_2 \)
- Rearranging gives:
\[ P_2 = P_0 \left(\frac{V_0}{\frac{V_0}{4}}\right)^{\frac{5}{3}} = P_0 \left(4\right)^{\frac{5}{3}} = P_0 \cdot 4^{\frac{5}{3}} = P_0 \cdot 8 \]
- Thus, \( P_2 = 8P_0 \).

**Hint:** Calculate the new pressure using the derived relationship.

### Step 5: Calculate the new temperature \( T_2 \)
- The temperature can be calculated using the ideal gas law:
\[ T = \frac{PV}{nR} \]
- Initial temperature \( T_1 = \frac{P_0 V_0}{nR} \) and final temperature:
\[ T_2 = \frac{P_2 V_2}{nR} = \frac{8P_0 \cdot \frac{V_0}{4}}{nR} = 2 \cdot \frac{P_0 V_0}{nR} = 2T_1 \]

**Hint:** Use the ideal gas law to find the relationship between pressure, volume, and temperature.

### Step 6: Calculate the work done in the right part
- The work done in an adiabatic process can be calculated using:
\[ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]
- Substituting the values:
\[ W = \frac{8P_0 \cdot \frac{V_0}{4} - P_0 V_0}{\frac{5}{3} - 1} = \frac{2P_0 V_0 - P_0 V_0}{\frac{2}{3}} = \frac{P_0 V_0}{\frac{2}{3}} = -2P_0 V_0 \]

**Hint:** Use the work formula for adiabatic processes to find the work done.

### Conclusion
Based on the calculations:
1. The pressure in the right part becomes \( 8P_0 \) (Option 1 is correct).
2. The temperature in the right part becomes \( 2T_1 \) (Option 2 is correct).
3. The work done by the gas is \( -2PV \) (Option 4 is correct).

Thus, the correct options are 1, 2, and 4.