Answer
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Hint: According to the first law of thermodynamics, the amount of heat energy given to the system equals the summation of total change in internal energy and net work done by the system.
$\Delta H = \Delta U + \Delta W$
But in a cyclic process, the total change in the internal energy of the system is equal to zero.
Complete step-by-step answer:
We know that the heat energy given to the system is converted into the work done and also to increase the internal energy of the system, given by the formula
$\Delta H = \Delta U + \Delta W$
Where,
$\Delta H$= Amount of heat supplied
$\Delta U$= Change in internal energy
$\Delta W$= Net work done
In a cyclic process, the initial and the final state of the system are the same therefore the net change in the internal energy of the system is zero.
$\therefore $$\Delta H = \Delta U + \Delta W$
On putting the value of $\Delta U = 0$we get
$\Delta H = 0 + \Delta W$
$\Delta H = \Delta W$
Therefore the heat supplied to the system becomes equal to the net work done by the system on the surrounding.
Hence the correct option is B.
Note: The amount of heat energy supplied to the system is also equal to the area enclosed under the P-V graph in a thermodynamic cycle. If the cycle is clockwise then work done by the system and if it is counterclockwise then the work done on the system by the surrounding. The work done by the system is considered positive and work done on the system is considered negative.
$\Delta H = \Delta U + \Delta W$
But in a cyclic process, the total change in the internal energy of the system is equal to zero.
Complete step-by-step answer:
We know that the heat energy given to the system is converted into the work done and also to increase the internal energy of the system, given by the formula
$\Delta H = \Delta U + \Delta W$
Where,
$\Delta H$= Amount of heat supplied
$\Delta U$= Change in internal energy
$\Delta W$= Net work done
In a cyclic process, the initial and the final state of the system are the same therefore the net change in the internal energy of the system is zero.
$\therefore $$\Delta H = \Delta U + \Delta W$
On putting the value of $\Delta U = 0$we get
$\Delta H = 0 + \Delta W$
$\Delta H = \Delta W$
Therefore the heat supplied to the system becomes equal to the net work done by the system on the surrounding.
Hence the correct option is B.
Note: The amount of heat energy supplied to the system is also equal to the area enclosed under the P-V graph in a thermodynamic cycle. If the cycle is clockwise then work done by the system and if it is counterclockwise then the work done on the system by the surrounding. The work done by the system is considered positive and work done on the system is considered negative.
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