Question

Two moles of an ideal gas undergoes a cyclic process as shown in the figure. If AB is an isothermal process at 223 $^\circ$C, then the net work done in the complete cycle is nearly: (Universal gas constant = 8.3 J mol$^{-1}$ K$^{-1}$ and take ln(2) = 0.7)

• A. 3810 J
• B. 5810 J
• C. 7810 J
• D. 2000 J

Hint:

For a cyclic process, calculate work done in each segment. Isothermal work involves $nRT \ln\left(\frac{V_2}{V_1}\right)$, isobaric involves $P \Delta V$, and isochoric work is zero.

Answer 1

The Correct Option is A

The cycle has three processes: AB (isothermal), BC (isobaric), and CA (isochoric). Net work done is the area of the cycle on the P-V diagram.
Step 1: Isothermal process AB (at T = 223°C = 496 K, from V = 2 m³ to 4 m³, P from 2000 N/m² to 1000 N/m²).
Work done W_AB = nRT ln(V_B / V_A) = 2 × 8.3 × 496 × ln(2) = 2 × 8.3 × 496 × 0.7 ≈ 5760 J.
Step 2: Isobaric process BC (from P = 1000 N/m², V = 4 m³ to V = 2 m³).
Work done W_BC = P ΔV = 1000 × (2 - 4) = -2000 J.
Step 3: Isochoric process CA (no work done, W_CA = 0).
Total work done = W_AB + W_BC + W_CA = 5760 - 2000 + 0 = 3760 J.
The closest option is 3810 J, likely due to rounding of ln(2) or R.