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Question: The heat absorbed by the system in going through the cyclic process as sown in figure is <img src="...

The heat absorbed by the system in going through the cyclic process as sown in figure is

A

30.4 J

B

31.4 J

C

32.4 J

D

33.4 J

Answer

31.4 J

Explanation

Solution

: In a cyclic process, heat absorbed by the system is equal to the work done by the system.

Heat absorbed = work done = Area of the given circle

=πr2=π×(ΔP2)(ΔV2)= \pi r ^ { 2 } = \pi \times \left( \frac { \Delta P } { 2 } \right) \left( \frac { \Delta V } { 2 } \right)

=3.14×(2002)×103×(2002)×10−6= 3.14 \times \left( \frac { 200 } { 2 } \right) \times 10 ^ { 3 } \times \left( \frac { 200 } { 2 } \right) \times 10 ^ { - 6 }

=3.14×100×103×100×10−6=31.4 J= 3.14 \times 100 \times 10 ^ { 3 } \times 100 \times 10 ^ { - 6 } = 31.4 \mathrm {~J}