Four moles of an ideal gas is initially in state `A` having pressure ` 2 xx 10^(5) N//m^(2)` and temperature ` 200 K` . Keeping the 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 equal to its volume at state `A`. Now gas is taken to initial state `A` keeping volume constant. A total `1000 J` of heat is withdrawn from the sample of the cyclic process . Take `R=8.3 J// K// mol`. The volume of gas at state `C` is

Options

A. `0.0332 m ^(3)`
B. `0.22 m^(3)`
C. `0.332m ^(3)`
D. `3.32 m ^(3)`

Explanation

To find the volume of the gas at state C, we can follow these steps:

### Step 1: Identify the Ideal Gas Law
The ideal gas law is given by the formula:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature

### Step 2: Calculate the Volume at State A
We know the following values for state A:
- Number of moles, \( n = 4 \) moles
- Pressure, \( P_A = 2 \times 10^5 \, \text{N/m}^2 \)
- Temperature, \( T_A = 200 \, \text{K} \)

Using the ideal gas law, we can rearrange the formula to solve for volume \( V \):
\[ V_A = \frac{nRT_A}{P_A} \]

Substituting the known values:
\[ V_A = \frac{4 \times 8.3 \, \text{J/(K mol)} \times 200 \, \text{K}}{2 \times 10^5 \, \text{N/m}^2} \]

### Step 3: Perform the Calculation
Calculating the numerator:
\[ 4 \times 8.3 \times 200 = 6640 \, \text{J} \]

Now, substituting this back into the volume equation:
\[ V_A = \frac{6640}{2 \times 10^5} = \frac{6640}{200000} = 0.0332 \, \text{m}^3 \]

### Step 4: Conclude the Volume at State C
Since it is given that the volume at state C is equal to the volume at state A:
\[ V_C = V_A = 0.0332 \, \text{m}^3 \]

Thus, the volume of gas at state C is:
\[ \boxed{0.0332 \, \text{m}^3} \]

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