Laws Of Thermodynamics

The problem involves calculating the total work done by the gas during a process path from state D to E and then to F. To solve this, we need to understand the processes involved:

1. Process D to E: This is a linear process where both pressure and volume change. The work done in a process is given by the area under the process curve in a PV-diagram.
2. Process E to F: This is an isobaric process (constant pressure), and the work done can be easily calculated using the formula W = P \times \Delta V .

Let's break this down step by step:

1. Calculate Work Done from D to E:
   • From the graph (approximation based on image understanding), the path DE is a straight line connecting D (600 N/m², 2 m³) to E (300 N/m², 5 m³).
   • The work done in the linear process can be approximated as the area under the line DE, which forms a trapezoid. The formula for the area of a trapezoid is:
     \text{Area} = \frac{1}{2} (b_1 + b_2) \times h
   • Here: b_1 = 600 \, \text{N/m}^2 , b_2 = 300 \, \text{N/m}^2 , and h = 5 - 2 = 3 \, \text{m}^3
   • Thus, the work done from D to E is:
     W_{DE} = \frac{1}{2} (600 + 300) \times 3 = \frac{1}{2} \times 900 \times 3 = 1350 \, \text{J}
2. Calculate Work Done from E to F:
   • This is an isobaric process at pressure 300 N/m².
   • The volume change from E to F is: 5 \, \text{m}^3 - 2 \, \text{m}^3 = -3 \, \text{m}^3 (indicating contraction).
   • The work done during E to F is:
     W_{EF} = 300 \times (-3) = -900 \, \text{J}
3. Total Work Done:
   • The total work done from D to E to F is given by the sum of the work done in each process:
   • W_{\text{total}} = W_{DE} + W_{EF} = 1350 - 900 = 450 \, \text{J}

Hence, the total work done by the gas is 450 J which matches option B.