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:
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The work done in thermodynamic processes can be calculated by finding the area under the curve in a \( P-V \) diagram. For a linear process, use the trapezoidal area, and for isobaric processes, the rectangular area.
Step 1: In the linear process from D to E, the work done is given by the area under the curve, which is a trapezoid. The work done in a trapezoidal process is: \[ W_{DE} = \frac{1}{2} \times (V_2 - V_1) \times (P_2 + P_1) \] where \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( P_1 \) and \( P_2 \) are the initial and final pressures. Step 2: From E to F, the volume is reduced at constant pressure, so the work done is: \[ W_{EF} = P \times \Delta V \] where \( P \) is the pressure during the isobaric process and \( \Delta V \) is the change in volume. Step 3: The total work done by the gas from D to E to F is the sum of the work done during both processes: \[ W_{{total}} = W_{DE} + W_{EF} \] After performing the calculations, we find that the total work done is 450J.
