Question

A thermodynamic system is taken from an original state D to an intermediate state E by the linear process shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

• A. -450 J
• B. 450 J
• C. 900 J
• D. 1350 J

Answer 1

The Correct Option is B

 To determine the total work done by the gas from state D to E to F, we need to analyze the processes described:

1. Process D to E: This is a linear process, which can be represented as a straight line on the PV diagram. The work done is the area under the line between these points. Since it's a straight line, the work can be calculated as the area of a trapezoid.
   • Initial pressure at D, \(P_D = 600 \, \text{N/m}^2\)
   • Final pressure at E, \(P_E = 300 \, \text{N/m}^2\)
   • Change in volume, \(\Delta V = 5 - 2 = 3 \, \text{m}^3\)
2. Process E to F: This is an isobaric process, meaning the pressure remains constant at 300 N/m².
   • Pressure, \(P_E = 300 \, \text{N/m}^2\)
   • Change in volume, \(\Delta V = 2 - 5 = -3 \, \text{m}^3\) (since volume is reducing)

To find the total work done from D to E to F:

\(W_{\text{total}} = W_{DE} + W_{EF} = 1350 - 900 = 450 \, \text{J}\)

Thus, the total work done by the gas is 450 J.