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 :

Solution

Physics sir · 4 steps

Work is area under the P–V curve: expansion A to B adds area, compression B to C takes some back.

1. A to B. The gas expands from 3 m³ to 7 m³ along a straight line, so the work done by it is the trapezium under that line. That is (8000 + 4000)/2 × 4 = 24000 dyne cm⁻² m³, and it is positive.
2. B to C. At constant pressure 4000 dyne cm⁻² the volume falls by 4 m³. The work done by the gas is -4000 × 4 = -16000 dyne cm⁻² m³, negative because the gas is compressed.
3. Fix the units. 1 dyne cm⁻² = (10⁻⁵ N)/(10⁻⁴ m²) = 0.1 Pa. So multiply by 0.1 to get joules.

   W = (24000 - 16000) × 0.1 J = 800 J

4. So. The net work equals the triangle ABC, 1/2 × 4 × 4000 × 0.1 = 800 J. The answer is (a).

The idea to keep. Area under the P–V path is the work. Expansion counts positive and compression counts negative. Before you multiply, convert pressure to pascals: 1 dyne cm⁻² = 0.1 Pa.

Answer from the NTA final answer key. Solution by PrepWiser.