Solveeit Logo
Login

Question

Question: Calculate $W_{BC}$ for 1mol ideal gas...

Calculate WBCW_{BC} for 1mol ideal gas

Answer

-800 cal

Explanation

Solution

To calculate WBCW_{BC} for 1 mol of an ideal gas:

  1. For cyclic process: ΔUTotal=0\Delta U_{Total} = 0.
  2. First Law of Thermodynamics: ΔU=Q+W\Delta U = Q + W (where W is work done on the system).
  3. WTotal=−QTotal=−600W_{Total} = -Q_{Total} = -600 cal.
  4. WTotal=WAB+WBC+WCAW_{Total} = W_{AB} + W_{BC} + W_{CA}.
  5. Process AB is isochoric, so WAB=0W_{AB} = 0.
  6. Process CA is isobaric, so WCA=−nRΔTCA=−1×2×(200−300)=−1×2×(−100)=200W_{CA} = -nR\Delta T_{CA} = -1 \times 2 \times (200 - 300) = -1 \times 2 \times (-100) = 200 cal.
  7. Substitute values: −600=0+WBC+200-600 = 0 + W_{BC} + 200.
  8. Solve for WBCW_{BC}: WBC=−600−200=−800W_{BC} = -600 - 200 = -800 cal.