An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas `(U=1.5 nRT)` and the part on the right contains two moles of the same gas. initially, the pressure on each side is p. the system is left for sufficient time so that a steady state is reached. find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c ) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.

Explanation

To solve the problem step by step, we will analyze each part of the question systematically.

### Given Data:
- Total volume \( V \) divided into two equal parts: \( V_1 = V_2 = \frac{V}{2} \)
- Left part contains 1 mole of ideal gas, right part contains 2 moles of the same gas.
- Initial pressure on both sides is \( P \).
- Internal energy of the gas is given as \( U = 1.5 nRT \).

### (a) Work done by the gas in the left part during the process
Since the separator is fixed, there is no change in volume for the gas in the left part. Therefore, the work done \( W \) can be calculated as:
\[ W = P \Delta V \]
Here, \( \Delta V = 0 \) (no change in volume), so:
\[ W = P \cdot 0 = 0 \]
Thus, the work done by the gas in the left part is **0**.

### (b) Temperature on the two sides in the beginning
Using the ideal gas law \( PV = nRT \), we can find the initial temperatures for both sides.

For the left part:
\[ PV_1 = n_1RT_1 \implies P \cdot \frac{V}{2} = 1 \cdot R \cdot T_1 \implies T_1 = \frac{PV}{2R} \]

For the right part:
\[ PV_2 = n_2RT_2 \implies P \cdot \frac{V}{2} = 2 \cdot R \cdot T_2 \implies T_2 = \frac{PV}{4R} \]

### (c) Final common temperature reached by the gases
The total internal energy before the process can be calculated as:
\[ U_{total} = U_{left} + U_{right} = n_1 C_v T_1 + n_2 C_v T_2 \]
Using \( C_v = \frac{3R}{2} \) for an ideal gas:
\[ U_{left} = 1 \cdot \frac{3R}{2} \cdot T_1 = \frac{3R}{2} \cdot \frac{PV}{2R} = \frac{3PV}{4} \]
\[ U_{right} = 2 \cdot \frac{3R}{2} \cdot T_2 = 3R \cdot \frac{PV}{4R} = \frac{3PV}{4} \]
Thus,
\[ U_{total} = \frac{3PV}{4} + \frac{3PV}{4} = \frac{3PV}{2} \]
The final temperature \( T_f \) can be calculated using:
\[ U_{total} = n_{total} C_v T_f \implies \frac{3PV}{2} = 3 \cdot \frac{3R}{2} \cdot T_f \]
Solving for \( T_f \):
\[ T_f = \frac{PV}{3R} \]

### (d) Heat given to the gas in the right part
Using the first law of thermodynamics:
\[ dQ = dU + dW \]
Since \( dW = 0 \):
\[ dQ = dU \]
The change in internal energy for the right part is:
\[ dU_{right} = n_2 C_v (T_f - T_2) = 2 \cdot \frac{3R}{2} \left( \frac{PV}{3R} - \frac{PV}{4R} \right) \]
Calculating:
\[ = 3 \left( \frac{PV}{3R} - \frac{PV}{4R} \right) = 3 \left( \frac{4PV - 3PV}{12R} \right) = \frac{PV}{4} \]

### (e) Increase in the internal energy of the gas in the left part
The change in internal energy for the left part is:
\[ dU_{left} = n_1 C_v (T_f - T_1) = 1 \cdot \frac{3R}{2} \left( \frac{PV}{3R} - \frac{PV}{2R} \right) \]
Calculating:
\[ = \frac{3R}{2} \left( \frac{2PV - 3PV}{6R} \right) = \frac{3R}{2} \cdot \left( -\frac{PV}{6R} \right) = -\frac{PV}{4} \]

### Summary of Answers:
(a) Work done by the gas in the left part: **0**
(b) Initial temperature on the left: \( T_1 = \frac{PV}{2R} \)
Initial temperature on the right: \( T_2 = \frac{PV}{4R} \)
(c) Final common temperature: \( T_f = \frac{PV}{3R} \)
(d) Heat given to the gas in the right part: \( \frac{PV}{4} \)
(e) Increase in the internal energy of the gas in the left part: \( -\frac{PV}{4} \)