Four moles of an ideal gas is initially in a state A having pressure `2xx10^(5)N//m^(2)` and temperature 200 K . Keeping pressure constant the gas is taken to state B at temperature of 400K. The gas is then taken to a state C in such a way that its temperature increases and volume decreases. Also from B to C, the magnitude of `(dT)/(dV)` increases. The volume of gas at state C is eaual to its volume at state A. Now gas is taken is initial state A keeping volume constant. A total of 1000 J heat is rejected from the sample in the cyclic process. Take `R=8.3J//K//mol`.
The work done by the gas along path B to C is

Options

A. `1000J`
B. `-1000J`
C. `-7640J`
D. `5640 J`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Process
We have an ideal gas undergoing a cyclic process:
- From state A to B (constant pressure, temperature increases from 200 K to 400 K).
- From state B to C (temperature increases, volume decreases).
- From state C back to A (volume constant).

### Step 2: Calculate Work Done in Each Segment
1. **From A to B**:
- Since the pressure is constant, the work done (W_AB) can be calculated using the formula:
\[ W_{AB} = P \Delta V \]
- We need to find the change in volume (ΔV). We can use the ideal gas law:
\[ PV = nRT \]
- At state A:
\[ V_A = \frac{nRT_A}{P} = \frac{4 \times 8.3 \times 200}{2 \times 10^5} = \frac{6640}{2 \times 10^5} = 0.0332 \, m^3 \]
- At state B (T_B = 400 K):
\[ V_B = \frac{nRT_B}{P} = \frac{4 \times 8.3 \times 400}{2 \times 10^5} = \frac{13280}{2 \times 10^5} = 0.0664 \, m^3 \]
- Thus, the change in volume from A to B is:
\[ \Delta V_{AB} = V_B - V_A = 0.0664 - 0.0332 = 0.0332 \, m^3 \]
- Work done from A to B:
\[ W_{AB} = P \Delta V_{AB} = 2 \times 10^5 \times 0.0332 = 6640 \, J \]

2. **From B to C**:
- We know that the total work done in the cyclic process is zero (since it returns to the initial state), and the heat rejected is -1000 J. Thus:
\[ Q_{total} = W_{AB} + W_{BC} + W_{AC} = 0 \]
- Rearranging gives us:
\[ W_{BC} = -1000 - W_{AB} - W_{AC} \]
- Since W_AC is zero (volume is constant), we have:
\[ W_{BC} = -1000 - 6640 = -7640 \, J \]

### Step 3: Conclusion
The work done by the gas along the path B to C is:
\[ W_{BC} = -7640 \, J \]