
In a cyclic process, the amount of heat given to a system is equal to
A. Net increase in internal energy
B. Net work done by the system
C. Net decrease in internal energy
D. Net change in volume
E. Net change in pressure
Answer
536.7k+ views
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.
Recently Updated Pages
Physics and Measurement Mock Test 2025 – Practice Questions & Answers

NCERT Solutions For Class 5 English Marigold - The Little Bully

NCERT Solutions For Class 12 Maths Three Dimensional Geometry Exercise 11.1

NCERT Solutions For Class 11 English Woven Words (Poem) - Ajamil And The Tigers

NCERT Solutions For Class 6 Hindi Durva - Bhaaloo

NCERT Solutions For Class 12 Physics In Hindi - Wave Optics

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
