Question

Thermodynamic change in the system from state \((P_{1},V_{1})\) to \((P_{2},V_{2})\) is shown in the figure. The work done by the system is

• A. \(16\times 10^{5}\ \text{J}\)
• B. \(12\times 10^{5}\ \text{J}\)
• C. \(8\times 10^{5}\ \text{J}\)
• D. \(4\times 10^{5}\ \text{J}\)

Hint:

Work done equals the area under the P-V graph between the initial and final volumes.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The work done by a gas equals the area under its P-V graph.

Step 2: Reading the graph
The path is a straight line from \(P=1\times10^{5}\ \text{N/m}^{2}\) at \(V=1\ \text{m}^{3}\) to \(P=3\times10^{5}\ \text{N/m}^{2}\) at \(V=5\ \text{m}^{3}\). The area below it is a trapezium. The volume increases, so the work is positive.

Step 3: Calculation
\(W=\dfrac{1}{2}(1+3)\times10^{5}\times(5-1)=8\times10^{5}\ \text{J}\)

Final Answer:
The work done by the system is 8 × 10^5 J. \[ \boxed{\text{(C) }8\times10^{5}\ \text{J}} \]