Question

A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be :

• A. 33800 J
• B. 2200 J
• C. 0 J
• D. 1200 J

Answer 1

The Correct Option is C

To determine the total work performed by the gas from state A to B, and subsequently from B to C, an analysis of the provided thermodynamic process is required, utilizing the principles of work calculation on a pressure-volume (P-V) diagram.

Work done by a gas during a process corresponds to the area delineated beneath the curve on a P-V diagram. Each path is examined individually:

1. Path A to B:
   • This path represents a linear process, indicating it is neither isothermal, isobaric, nor isochoric.
   • The calculation of work done from A to B necessitates the formula for the area under a straight line on a P-V diagram, typically computed using trapezoidal or triangular area methodologies.
2. Path B to C:
   • This process is isobaric, characterized by constant pressure.
   • The work done in an isobaric process is quantified by the formula: \(W = P \Delta V\), where \( P \) denotes the constant pressure and \( \Delta V \) represents the change in volume.
   • Observing the diagram, given that \( V_{\text{C}} = V_{\text{A}} \), there is no net change in volume for the overall process returning to the initial state, resulting in \(\Delta V = 0\).

As the volume reverts to its original value, any work performed from A to B is effectively negated by the work done from B to C, assuming ideal conditions without additional energy losses or gains. Consequently, the aggregate work conducted throughout the complete cycle from A to B and then to C is:

\(W_{\text{total}} = W_{\text{AB}} + W_{\text{BC}} = 0\)

Therefore, the definitive result is 0 J.