A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is
- 20J
- 30J
- -30J
- -60J
The Correct Option is B
Solution and Explanation
\(\Delta U_{AB} = 40 J\) as process is isochoric.
\(\Delta U_{BC} + W_{BC} = 0\)
\(\Delta U_{BC} = +50\) \((W_{BC} = –50 J)\)
\(U_C = U_A + \Delta U_{AB} + \Delta U_{BC} = 1650\)
For CA process,
\(Q_{CA} = – 60 J\)
\(\Delta U_{CA} + W_{CA} = –60\)
\(–90 + W_{CA} = –60\)
\(\Rightarrow W_{CA} = +30 J\)
The graph given is inconsistent with the statement BC may be adiabatic and CA cannot be like isobaric as shown, as increasing volume while rejecting heat at same time.
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Concepts Used:
Work-Energy Theorem
The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.
The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,
W = ΔKE = ½ mv2f − ½ mv2i
Where,
vi → Speeds of the particle before the application of force
vf → Speeds of the particle after the application of force
m → Particle’s mass
Note: Energy and Momentum are related by, E = p2 / 2m.