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

We need to find the total work done by the gas during the thermodynamic process from $$D \rightarrow E \rightarrow F$$.

The total work done ($$W_{\text{total}}$$) is the sum of the work done in each individual stage of the path:

$$W_{\text{total}} = W_{DE} + W_{EF}$$

Step 1: Work Done during the Linear Process ($$D \rightarrow E$$)

The work done during any process on a Pressure-Volume ($$P-V$$) diagram is equal to the area under the $$P-V$$ curve projected onto the Volume axis.

• The region under the straight line path $$DE$$ forms a trapezium bounded between $$V_D = 2.0\text{ m}^3$$ and $$V_E = 5.0\text{ m}^3$$.
• The formula for the area of a trapezium is: $$\text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Distance between them})$$
• Parallel sides (Pressures): $$P_D = 600\text{ N m}^{-2}$$ and $$P_E = 300\text{ N m}^{-2}$$
• Distance between them (Change in Volume): $$\Delta V = V_E - V_D = 5.0 - 2.0 = 3.0\text{ m}^3$$

Calculating $$W_{DE}$$:

$$W_{DE} = \frac{1}{2} \times (600 + 300) \times (5.0 - 2.0)$$

$$W_{DE} = \frac{1}{2} \times 900 \times 3.0 = 450 \times 3.0 = +1350\text{ J}$$

Note: The work is positive because the gas is expanding from $$2.0\text{ m}^3$$ to $$5.0\text{ m}^3$$.

Step 2: Work Done during the Isobaric Process ($$E \rightarrow F$$)

The process from $$E \rightarrow F$$ is an isobaric compression path carried out at a constant pressure.

• Constant Pressure ($$P$$): $$300\text{ N m}^{-2}$$
• Initial Volume ($$V_E$$): $$5.0\text{ m}^3$$
• Final Volume ($$V_F$$): $$2.0\text{ m}^3$$

Calculating $$W_{EF}$$:

$$W_{EF} = P \times \Delta V = P \times (V_F - V_E)$$

$$W_{EF} = 300 \times (2.0 - 5.0) = 300 \times (-3.0) = -900\text{ J}$$

Note: The work is negative because the gas is undergoing compression.

Step 3: Calculate Total Work Done

Combining the work computed from both segments gives:

$$W_{\text{total}} = W_{DE} + W_{EF}$$

$$W_{\text{total}} = 1350\text{ J} + (-900\text{ J}) = 450\text{ J}$$

Therefore, the total work done by the gas from D to E to F is 450 J, which corresponds to Option B.